Hc Verma II for Class 11 Science Physics Chapter 28 - Heat Transfer

Answer:

The amount of heat crossing through any cross-section of a slab in time $∆t$ is called heat current.
It is written as $\frac{∆\mathrm{Q}}{∆t}$ and not as complete derivative $\frac{d\mathrm{Q}}{dt}$. This is because the amount of heat crossing through any cross section is a function of many variables like temperature difference, area of cross-section, etc. So, we cannot write it as a complete derivative with respect to time.

Answer:

Yes, the body will radiate. However, its temperature will not fall down with time because as the temperature of the surroundings is greater than the temperature of the body so, its rate of absorption will be greater than its rate of emission.

Answer:

Here, major role is played by convection. When we blow air over a spoonful of hot tea, the air coming from our mouth has less temperature than the air above the tea. Since hot air has less density, it rises up and cool air goes down. In this way, the tea cools down.
We know that any hot body radiates. So, the spoonful of tea will also radiate and as the temperature of the surrounding is less then the tea, the tea will cool down with time. Evaporation is also involved in this. On blowing over the hot tea, rate of evaporation increases and the cools down.

Answer:

No. When the door of the refrigerator is left open in a closed room, the heat given out by the refrigerator to the room will be more than that taken from the room. Therefore, instead of decreasing, the temperature of the room will increase at a slower rate.

Answer:

We will prefer to seat on a wooden chair because as the conductivity of wood is poorer than that of metal, heat flow from our body to the chair will be less in case of a wooden chair.

Answer:

Yes, the temperature of the balls can be equalised by radiation. This is because both the spheres will emit radiations in all the directions at different rates.
The ball kept at the temperature of 300 K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600 K. Also, it losses energy by radiation.
Similarly, the ball kept at the temperature of 600 K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600 K. Also, it losses energy by radiation.
A time comes when the temperature of both the bodies becomes equal.

Yes, the rate of heat gained by the colder sphere is proportional to $T_2^4 - T_1^4$.

Answer:

An ordinary electric fan does not cool the air, still it gives comfort in summer because it circulates the air present in the room. Due to this, evaporation takes place and we feel cooler.

Answer:

The temperature of the atmosphere at a high altitude is around 500°C, but density of air molecule is extremely low at this height. So, very less molecules of air collide with the body of the animal and transfer very less amount of heat. That is why the animal present there would freeze to death instead boiling.

Answer:

The heat coming from the sun to us is through the radiation. On colder winter days, when we stand in shade, we do not get the heat of the sun from the radiation. Though we feel cool in the shade, the temperature of the air in shady as well as non-shady regions is the same.

Answer:

During night, the earth's surface radiates infrared radiation of larger wavelength. Gas molecules in the air absorb some of this energy and radiate energy of their own in all directions. Also, water molecules, like the vapour that makes the clouds, absorb more frequencies of infrared energy than clear air does.
Both these factors contribute to the fact that clouds radiate more heat in all directions (including the earth) than clear air does. In turn, this makes the overall temperature on the earth warmer when there is a cloud cover. The heat energy radiated by the earth is reflected back to earth. Due to this, cloudy nights are warmer than the nights with clean sky.

Answer:

A white colour dress reflects almost all the radiations falling on it. So, it does not absorb any heat from the sunlight and we feel more comfortable in it. On the other hand, a dark colour dress absorbs maximum radiation falling on it. So, we feel hot in a dark coloured dress during summers.

Answer:

(d) material of the rod

The thermal conductivity of a rod depends only on the material of the rod. For example, metals are much better conductors than non-metals because metals have large number of free electron that can move freely anywhere in the body of the metal and carry thermal energy from one place to other. Also, 2 copper rods having different lengths and areas of cross-section have same thermal conductivity that depends only on the number of free electrons in copper.

Answer:

(d) by all the three modes

In conduction, heat is transferred from one place to other by vibration of the molecules. In this process, the average position of a molecule does not change. Hence, there is no mass movement of matter.
In convection, heat is transferred from one place to other by actual motion of particles of the medium. When water is heated, hot water moves upwards and cool water moves downwards.
In radiation process, transfer of heat does not require any material medium.
For a room containing air, heat can be transferred via radiation (no medium required) and convection (by the movement of air molecules) and by conduction (due to collision of hot air molecules with other molecules).

Answer:

(d) $T_2^4 -\mathit{ }{T}_1^4$

From Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
    $u = \sigma A{T}^4$
Here, $\sigma$ is Stefan-Boltzmann constant.
Since the temperature of the solid is less than the surroundings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The rate of emission from the solid will be proportional to ${T_1}^4$ and rate of emission from the surroundings will be proportional to ${T_2}^4$.
So, the net rate of increase in temperature will be proportional to $T_2^4 - T_1^4$.

Answer:

(b) all bodies

From Stefan-Boltzmann law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
     $u = \sigma A{T}^4$
Here, $\sigma$ is Stefan-Boltzmann constant.
This law holds true for all the bodies.

Answer:

(a) 1 : 1.15

From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
     $u = \sigma A{T}^4$
Here, $\sigma$ is Stefan-Boltzmann constant.

The thermal radiation emitted in a given time by A and B will be in the ratio
$$\begin{gathered} \frac{u_{\mathrm{A}}}{u_{\mathrm{B}}} = \frac{{T_{\mathrm{A}}}^4}{{T_{\mathrm{B}}}^4} \\ \frac{u_{\mathrm{A}}}{u_{\mathrm{B}}}=\frac{(273+10{)}^4}{(273+20{)}^4} \\ \frac{u_{\mathrm{A}}}{u_{\mathrm{B}}}=\frac{1}{1.15} \end{gathered}$$

Answer:

(d) in nonuniform

In steady state, the temperature of the rod is nonuniform maximum at the end near the furnace and minimum at the end that is away from the furnace.

Answer:

(c) Stefan's law

From Stefan-Boltzman's law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
   $u = \sigma A{T}^4$
Where $\sigma$ is Stefan's constant.
Suppose a body at temperature T is kept in a room at temperature T₀.

According to Stefan's law, energy of the thermal radiation emitted by the body per unit time is
  $u = e\sigma A{T}^4$
Here, e is the emissivity of the body.

The energy absorbed per unit time by the body is (due to the radiation emitted by the walls of the room)
 $u_0 = e\sigma A{T_0}^4$

Thus, the net loss of thermal energy per unit time is
$$\begin{gathered} ∆u = u-u_0 \\ ∆u=e\sigma A(T^4-{T_0}^4) ...\left(\mathrm{i}\right) \end{gathered}$$

Newton law of cooling is given by
$\frac{\mathit{d}T}{\mathit{d}t} = -bA(T-T_0)$

This can be obtained from equation (i) by considering the temperature difference to be small and doing the binomial expansion.

Answer:

(a)
When a hot liquid is kept in a big room, the liquid will loose its temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to $T^4$, where T is the initial temperature of the liquid.
As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease.
Therefore, the plot of temperature with time is best represented by the curve in option (a).

Answer:

(a) a straight line

When a hot liquid is kept in a big room, then the liquid will loose temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to $T^4$, where T is the initial temperature of the liquid. As the temperature decreases, the rate of loss will also decrease. So, the slope of the curve will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid. So, in steady state, the difference between the temperatures of the two will become zero.

A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time.The logarithm converts the fourth power dependence into a linear dependence with some coefficient (property of log). So, the plot satisfying all the above properties will be a straight line.

Answer:

(c) more than 5 minutes

Let the temperature of the surrounding be T$°\mathrm{C}$.
Average temperature of the liquid in first case = 62.5$°\mathrm{C}$
Average temperature difference from the surroundings = (62.5 $-$ T)$°\mathrm{C}$

From newton law of cooling,

$$\begin{gathered} 1^{°}\mathrm{C} \min { }^{-1}=-bA(62.5-T{)}^{°}\mathrm{C} \\ \Rightarrow -bA\mathit{ }= \frac{1}{62.5-T}{\min }^{-1} ... (\mathrm{i}) \end{gathered}$$

For the second case,
Average temperature = 57.5$°\mathrm{C}$
Temperature difference from the surroundings = (57.5 $-$ T)$°\mathrm{C}$

From Newton's law of cooling and equation (i), 
$$\begin{gathered} \frac{5^{°}\mathrm{C}}{t}=-bA(57.5-T{)}^{°}\mathrm{C} \\ \Rightarrow \frac{5^{°}\mathrm{C}}{t}= \frac{1}{62.5-T}(57.5-T{)}^{°}\mathrm{C} \\ \Rightarrow t\mathit{ }=\frac{5( 62.5-T)}{(57.5-T)} \\ \Rightarrow t>5 \mathrm{minutes} \end{gathered}$$

Answer:

(d) The state of the rod does not change after steady state is reached

The heat transfer will take place from the hot end to the cold end of the rod via conduction. So, with time, the temperature of the rod will increase from the end dipped in boiling water to the end dipped in melting, until it comes in equilibrium with its surroundings. In steady state, the temperature of the rod is nonuniform and constant, maximum at the end dipped in boiling water and minimum at the end dipped in melting ice.
Equilibrium means that the system is stable. So, all the macroscopic variables describing the system will not change with time. Hence, the temperature of the rod will become constant once equilibrium is reached, but its value is different at different positions of the rod.

Answer:

(c) reflect radiation
(d) refract radiation

A black body is an ideal concept. A black body is the one that absorbs all the radiation incident on it. So, a black body does not reflect and refract radiation.

Answer:

(b) convection current

Convection current is the movement of air (or any fluid) due to the difference in the temperatures. During summer days, there is temperature difference of air above the land and river. Due to this, a convection current is set from the river to the land during daytime. On the other hand, during night, a convection current is set from the land to the river. Therefore, a mild air always flows on the shore of a calm river due to the convection current.  

Answer:

(c) If both are picked up by bare hands, the steel will be felt hotter than the charcoal
(d) If the two are picked up from the lawn and kept in a cold chamber, the charcoal will lose heat at a faster rate than the steel.

In steel, conductivity is higher than charcoal. So, if both are picked up by bare hands, then heat transfer from the body (steel or charcoal) to our hand will be larger in case of steel. Hence, steel will be hotter than the charcoal.
On the other hand, emissivity of charcoal is higher as compared to steel. So, if the two are picked up from the lawn and kept in a cold chamber, charcoal will lose heat at a faster rate than steel.

Answer:

(a) the maximum intensity of radiation will be near the frequency 2v₀
(c) the total energy emitted will increase by a factor of 16

From Wein's displacement law,
${\lambda }_{\mathrm{m}}T$ = b (a constant)

or $\frac{cT}{{\nu }_{\mathrm{m}}}=b$

Here, T is the absolute temperature of the body.

So, as the temperature is doubled to keep the product on the left hand side constant, frequency is also doubled.

From Stefan's law, we know that the rate of energy emission is proportional to $T^4$.
This implies that total energy emitted will increase by a factor of (2)⁴, which is equal to 16.

Answer:

(a) Both will emit equal amount of radiation per unit time in the beginning.
(b) Both will absorb equal amount of radiation from the surrounding in the beginning.

Let the temperature of the surroundings be $T_0$.
From the Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
$u = \sigma A{T}^4$
Here, $\sigma$ is Stephen's constant.
Also, the energy absorbed per unit time by the body is given by
$u_0 = e\sigma A{T_0}^4$
As the two spheres have equal radii and temperatures, their rate of absorption and emission will be equal in the beginning.

Answer:

Given:
Thermal conductivity of the material, k = 0.80 W m^–1 °C^–1
Area of the cross section of the slab, A = 100 cm² = 10⁻² m²
Thickness of the slab, Δx = 1 cm = 10⁻² m

$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}=\frac{\mathrm{Temperature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{\Delta T}{{\displaystyle \frac{\mathrm{\Delta }x}{k·A}}} \\ \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{\left(90-10\right) k·\mathrm{A}}{\mathrm{\Delta }x} \\ \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{\left(80\right)\times 0.8\times {10}^{-2}}{{10}^{-2}} \\ \frac{\Delta Q}{\mathrm{\Delta }t}=64 \mathrm{J}/\mathrm{s} \\ \frac{\Delta Q}{\mathrm{\Delta }t}=64\times 60=3840 \mathrm{J}/\min \end{gathered}$$

Answer:

$\mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}=\frac{\mathrm{Temperature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}}$
Thickness of the container, l = 1 cm = 10^$-$2 m
Thermal conductivity of the styrofoam sheet, $k=0.025 \mathrm{J} {\mathrm{s}}^{-1} {\mathrm{m}}^{-1} °{\mathrm{C}}^{-1}$
Area, A = 0.80 m²
Thermal resistance, $\frac{l}{k\mathit{ }A}=\frac{{10}^{-2}}{{\displaystyle 0.025\times 0.80}}$
Temperature difference, $∆T=T_1-T_2=300-80=220 \mathrm{K}$

Rate of flow of heat, $\left(\frac{\Delta Q}{\Delta t}\right)=\frac{T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{kA}}}$
$$\begin{gathered} \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=\frac{220}{{10}^{-2}}\times 0.025\times 0.80 \\ \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=440 \mathrm{J}/s \end{gathered}$$

Answer:

$\mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}=\frac{\mathrm{Temperature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}}$
Temperature of the body, $T_1=97°\mathrm{F}=36.11°\mathrm{C}$
Temperature of the surroundings, $T_2=47°\mathrm{F}=8.33°\mathrm{C}$
Conductivity of the cloth, $K=0.04 \mathrm{J} {\mathrm{s}}^{-1} {\mathrm{m}}^{-1} °{\mathrm{C}}^{-1}$
Thickness of the cloth, $l=0.5 \mathrm{cm}=0.005 \mathrm{m}$
Area of the cloth, $A=1.6 {\mathrm{m}}^2$

Difference in the temperature, $∆T=T_1-T_2=36.11-8.33=27.78°\mathrm{C}$
Thermal resistance = $\frac{l}{KA}=\frac{0.005}{\left(0.04\right)\times 1.6}$ = 0.078125

Rate at which heat is flowing out is given by

$$\begin{gathered} \frac{\Delta Q}{\Delta t}=\frac{T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{KA}}} \\ \frac{\Delta Q}{\Delta t}=\frac{27.78}{0.078125} \\ \frac{\Delta Q}{\Delta t}=356 \mathrm{J}/s \end{gathered}$$
 

Answer:

Area of the bottom of the container, $A=25 {\mathrm{cm}}^2=25\times {10}^{-4} {\mathrm{m}}^2$
Thickness of the bottom of the container, l = 1 mm = 10^$-$3 m
Latent heat of vaporisation of water, $L=2.26\times {10}^6 \mathrm{J} {\mathrm{kg}}^{-1}$
Thermal conductivity of the container, $k=50 \mathrm{W} {\mathrm{m}}^{-1}°{\mathrm{C}}^{-1}$
mass = 100 g = 0.1 kg

Rate of heat transfer from the base of the container is given by
$$\begin{gathered} \frac{\Delta Q}{\Delta t}=\frac{m L}{\Delta t}=\frac{\left(0.1\right)\times 2.26\times {10}^6}{1 \min } \\ \frac{\Delta Q}{\Delta t}=0.376\times {10}^4 \mathrm{J}/\mathrm{s} \end{gathered}$$
Also,
$$\begin{gathered} \frac{\Delta Q}{\Delta t}=\frac{\Delta T}{{\displaystyle \frac{l}{kA}}} \\ \Rightarrow 0.376\times {10}^4=\frac{T-100}{{\displaystyle \frac{{10}^{-3}}{50\times 25\times 10{ }^{-4}}}} \\ \Rightarrow 0.376\times {10}^4=\frac{50\times 25\times {10}^{-4} \left(T-100\right)}{{10}^{-3}} \\ \Rightarrow \left(T-100\right)=3.008\times 10 \\ \Rightarrow T=130°\mathrm{C} \end{gathered}$$

Answer:

Given:
Thermal conductivity of the rod, K = 46 J s^–1 m^–1 °C^–1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A = 0.04 cm²
                                                       = 0.04 × 10⁻⁴ m²
                                                        = 4 × 10⁻⁶ m²
Rate of transfer of heat is given by
$$\begin{gathered} \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\Delta T}{{\displaystyle \frac{l}{\mathrm{kA}}}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\left(T_1-T_2\right) kA}{l} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\left(\frac{100-0}{1}\right)\times 46\times 4\times {10}^{–6} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=184\times {10}^{-4 }\mathrm{J}/\mathrm{s} \end{gathered}$$


$$\begin{gathered} \mathrm{Also}, \mathrm{\Delta Q}=m{\mathrm{L}}_f \\ \therefore \left(\frac{m}{t}\right)L_f=184\times {10}^{-4} \\ \Rightarrow \left(\frac{m}{t}\right)\times 3.36\times {10}^5=184\times {10}^{–4} \\ \Rightarrow m=\frac{184\times {10}^{-4}}{3.36\times {10}^5}\times t \\ \Rightarrow m=5.5\times 10\times {10}^{-9}\times 1 \mathrm{kg}/\mathrm{s} \\ \Rightarrow m=5.5\times {10}^{-8}\times {10}^3 \mathrm{g}/\mathrm{s} \\ \Rightarrow m=5.5\times {10}^{-5} \mathrm{g}/\mathrm{s} \end{gathered}$$

Answer:

Area of the walls of the box, A = 2400 cm² = 2400 × 10⁻⁴ m²
Thickness of the ice box, l = 2 mm = 2 × 10⁻³ m
Thermal conductivity of the material of the box, K = 0.06 W m⁻¹ °C⁻¹
Temperature of the water outside the box, T₁ = 20°C
Temperature of ice, T₂ = 0°C

$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat} =\frac{\mathrm{Temperature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{T_1-T_2}{{\displaystyle \frac{l}{kA}}} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\left(\frac{20}{2\times {10}^{-3}}\right)\times 0.06\times 2400\times {10}^{-4} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=24\times 6=144 \mathrm{J}/\mathrm{s} \end{gathered}$$
Rate at which the ice melts =$\frac{m{L}_f}{t}$
$$\begin{gathered} \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\left(\frac{m}{t}\right)L_f \\ \Rightarrow 144=\left(\frac{m}{t}\right)\times 3.4\times {10}^5 \\ \Rightarrow \frac{m}{t}=\frac{144}{3.4\times {10}^5} \mathrm{kg}/\mathrm{s} \\ \Rightarrow \frac{m}{t}=\frac{144\times 60\times 60}{3.4\times {10}^5} \mathrm{kg}/\mathrm{h} \\ \Rightarrow \frac{m}{t}=1.52 \mathrm{kg}/\mathrm{h} \end{gathered}$$

Answer:

Thickness of porous walls, l = 1 mm = 10⁻³ m
Mass, m = 10 kg
Latent heat of vapourisation, L_v = 2.27 × 10⁶ J/kg
Thermal conductivity, K = 0.80 J/m s °C
ΔQ = 2.27 × 10⁶ × 10 J
0.1 g of water evaporate in 1 sec, so 10 kg water will evaporate in 10⁵ s.

$$\begin{gathered} \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{2.27\times 10{ }^7}{{10}^5} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=2.27\times {10}^2 \mathrm{J}/\mathrm{s} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{\Delta T}{{\displaystyle \frac{l}{kA}}} \\ \Rightarrow \frac{\Delta \mathrm{Q}}{\mathrm{\Delta }t}= \left(\frac{42-T}{{10}^{-3}}\right)·0.80\times 2\times {10}^{-2} \\ \Rightarrow 2.27\times {10}^2=\frac{\left(42-\mathrm{T}\right)}{{10}^{-3}}\times 0.80\times 2\times {10}^{-2} \\ \Rightarrow T= 27.8°\mathrm{C} \\ \Rightarrow T=28°\mathrm{C} \end{gathered}$$

Answer:

Thermal conductivity, K = 45 W m^–1 °C^–1
Length, l = 60 cm = 0.6 m
Area of cross section, A = 0.2 cm² = 0.2 × 10⁻⁴ m²
Initial temperature, T₁ = 40°C
Final temperature, T₂ = 20°C
$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}=\frac{\mathrm{Temerature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{kA(T_1-T_2)}{{\displaystyle l}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{45\times 0.2\times {10}^{-4} \left(40 - 20\right)}{0.6} \\ =0.03 \mathrm{W} \end{gathered}$$

Answer:

Area of cross section, A = 10 cm² = 10 × 10^–4 m² 
Thermal conductivity, K = 200 Js^–1 m^–1 °C^–1
Height, H = 10 cm
Length, l = 1 mm =10^–3 m

Temperature inside the cylindrical vessel, T₁ = 50°C
Temperature outside the vessel, T₂ = 30°C



Rate of flow of heat from 1 flat surface will be given by

$$\begin{gathered} \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{T_1-T_2}{{\displaystyle \frac{l}{\mathrm{kA}}}} \\ \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{ \left(50-30\right)\times 200\times {10}^{-3}}{{10}^{-3}} \\ \frac{\Delta Q}{\mathrm{\Delta }t}=6000 \mathrm{J}/\sec \end{gathered}$$

Heat escapes out from both the flat surfaces.
Net rate of heat flow = 2 × 6000 = 12000 J/sec

$\frac{\Delta Q}{\mathrm{\Delta }t}=\frac{m·\mathrm{s}·\Delta T}{{\displaystyle \mathrm{\Delta }t}}$

Mass = Volume density
⇒ $$\begin{gathered} {10}^{-3}\times 0.1\times 1000 \\ =0.1 \mathrm{g} \end{gathered}$$

Using this in the above formula for finding the rate of flow of heat, we get

 $$\begin{gathered} 12000=0.1\times 4200\times \left(\frac{\Delta T}{t}\right) \\ \Rightarrow \frac{\mathrm{\Delta }T}{\mathrm{\Delta }t}=\frac{ 12000}{420}=28.57 \\ \mathrm{As} \mathrm{\Delta }T=1°\mathrm{C} \\ \Rightarrow \frac{1}{t}=28.57 \\ \Rightarrow t=\frac{1}{28.57}= 0.035 \sec \end{gathered}$$

Answer:

Area of cross section, A = 0.2 cm² = 0.2 × 10⁻⁴ m²
Thermal conductivity, k = 385 W m^–1 °C^–1

 $$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}=\frac{\mathrm{Temperature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t} = \frac{kA\mathit{(}{T}_1-T_2\mathit{)}}{{\displaystyle l}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t} = \left(\frac{80-20}{0.2}\right)\times 385\times 0.2\times {10}^{-4} \\ = 2310\times {10}^{-3} \\ = 2.31 \mathrm{J}/\sec \end{gathered}$$

Let te temperature of point C be T, which is at a distance of 11 cm from the left end.
Rate of flow of heat is given by

 $$\begin{gathered} \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t} = \frac{\mathrm{k}A\mathrm{\Delta }T}{{\displaystyle l}} \\ \Rightarrow \frac{\mathrm{\Delta }T}{l} = \left(\frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}\right)\times \frac{1}{\mathrm{kA}} \\ \frac{T-20}{11\times {10}^{-2}} = \frac{2.31}{383\times 0.2\times {10}^{-4}} \\ \Rightarrow T = 33+20 \\ \Rightarrow T = 53°\mathrm{C} \end{gathered}$$

Answer:

One end of the rod is at a temperature of 25°C. So, if no heat current flows through the rod in steady state, then the other end of the rod should also be at a temperature of 25°C.
Let the point at which the other end of the rod is touched be C.
No heat flows through the rod when the temperature at point C is also 25°C.

Heat current through AC = Heat current through CB

 $$\begin{gathered} \Rightarrow \frac{{\left(\mathrm{\Delta }T\right)}_{\mathrm{AC}}}{{\displaystyle \frac{x}{\mathrm{kA}}}} = \frac{{\left(\mathrm{\Delta }T\right)}_{\mathrm{CB}}}{{\displaystyle \frac{100-x}{\mathrm{kA}}}} \\ \frac{\left(100-25\right)}{x} = \frac{25-0}{100-x} \\ \frac{3}{x} = \frac{1}{100-x} \\ 300-3x = x \\ 300 = 4x \\ x = 75 \mathrm{cm} \end{gathered}$$

Thus, it should be touched at 75 cm from 100°C end.

Answer:

Volume of the cube, V = a³ = 216 cm³
Edge of the cube, a = 6 cm
Surface area of the cube = 6a²
                                  = 6 (6 × 10⁻²)²
                                  = 216 × 10⁻⁴ m²
Thickness, l = 0.1 cm = 0.1 × 10^–2 m
Temperature difference, Δ T = 5°C
The inner surface of the cube is heated by a 100 W heater.
∴ Power, P = 100 W
Power = Energy per unit time
∴ Rate of flow of heat inside the cube, R = 100 J/s
Rate of flow of heat is given by
 $$\begin{gathered} \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\Delta \mathrm{T}}{{\displaystyle \frac{l}{\mathrm{kA}}}} \\ 100=\frac{\mathrm{k}\times 216\times {10}^{-4}\times 5}{0.1\times {10}^{-2}} \\ \mathrm{k}=0.9259 \mathrm{W}/\mathrm{m}°\mathrm{C} \end{gathered}$$

Answer:

Temperature of water, T₁ = 1°C
Temperature if ice bath, T₂ = 0°C
Thermal conductivity, K = 0.5 W/m °C
Length through which heat is lost, l = 2 mm = 2 × 10^–3 m
Area of cross section, A = 5 × 10⁻² m²
Velocity of the block, v = 10 cm/sec = 0.1 m/s
Let the mass of the block be m.
Power = F · v
           = (mg) v          ......(i)
Also,
 $$\begin{gathered} \mathrm{Power}=\frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t} .......\left(\mathrm{ii}\right) \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }\mathrm{t}} = \frac{\mathrm{k}·\mathrm{A} \left({\mathrm{T}}_1-{\mathrm{T}}_2\right)}{\mathrm{l}} ........\left(\mathrm{iii}\right) \end{gathered}$$

From equation (i), (ii) and (iii), we get
$$\begin{gathered} \left(\mathit{m}\mathit{g}\right)\mathit{ }v=\frac{\mathrm{k}·\mathrm{A} \left({\mathrm{T}}_1-{\mathrm{T}}_2\right)}{l} \\ m=\frac{0.5\times 5\times {10}^{-2} \left(1\right)}{\left(2\times {10}^{-3}\right)\times 10\times 0.1} \\ m=12.5 \mathrm{kg} \end{gathered}$$

Answer:

Thermal conductivity, K = 1.7 W/m°C
Density of water, ρ_ω = 10² kg/m³
Latent heat of fusion of ice, L_ice = 3.36 × 10⁵ J/kg
Length, l = 10 × 10⁻² m

(a) Rate of flow of heat is given by
$\frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)·\mathrm{KA}}{l}$
$$\begin{gathered} \Rightarrow \frac{l}{\mathrm{\Delta }t}=\frac{\left(T_1-T_2\right)\mathit{·}KA}{\mathrm{\Delta Q}} \\ =\frac{\mathrm{K}·\mathrm{A} \left({\mathrm{T}}_1-{\mathrm{T}}_2\right)}{\mathrm{m}·\mathrm{L}} \\ = \frac{\mathrm{KA}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)}{\left(\mathrm{A}l·{\mathrm{\rho }}_{\mathrm{\omega }}\right) \mathrm{L}} \\ = \frac{\left(1.7\right) \left(0-10\right)}{\left(10\times {10}^{-2}\right)\times {10}^3\times 3.36\times {10}^5} \\ =\frac{17}{3.36}\times {10}^{-7} \\ =5.059\times {10}^{-7} \mathrm{m}/\mathrm{s} \\ =5\times {10}^{-7} \mathrm{m}/\mathrm{s} \end{gathered}$$

(b) To form a thin ice layer of thickness dx, let the required be dt.
Mass of that thin layer, dm = A dx ρ_ω
Heat absorbed by that thin layer, dQ = Ldm

$$\begin{gathered} \frac{d\mathrm{Q}}{dt} = \frac{K·A \left(∆T\right)}{x} \\ \frac{L\mathrm{d}m}{\mathrm{d}t} = \frac{KA\mathit{ }\left(∆T\right)}{x} \\ \frac{\left(\mathrm{A}dx{\mathrm{\rho }}_w\right)L}{\mathrm{d}t} = \frac{KA\mathit{∆}T}{x} \\ \Rightarrow \underset{0}{\overset{t}{\int }}\mathrm{d}t = \frac{{\rho }_{w}L}{K\mathit{ }\left(\Delta T\right)} \underset{0}{\overset{0.1}{\int }}x dx \\ \Rightarrow t = \frac{{\rho }_{w}L}{K \left(\Delta T\right)} {\left[\frac{x^2}{2}\right]}_0^{0.1} \\ t = \frac{{\rho }_{w}L}{K\mathit{ }\left(\Delta T\right)}\times \left(\frac{{\left(0.1\right)}^2}{2}\right) \\ t = \frac{{10}^3\times 3.36\times {10}^5\times 0.01}{1.7\times 10\times 2} \\ t = \frac{3.36}{2\times 17}\times {10}^6 \sec \\ t = \frac{3.36}{2\times 17}\times \frac{{10}^6}{3600} \mathrm{hours} \\ t = 27.45 \mathrm{hours} \end{gathered}$$

Answer:

Let the point upto which ice is formed is at a distance of x m from the top of the lake.
Under steady state, the rate of flow of heat from ice to this point should be equal to the rate flow of heat from water to this point.
Temperature of the top layer of ice = −10°C
Temperature of water at the bottom of the lake = 4°C
Temperature at the point upto which ice is formed = 0°C
$$\begin{gathered} {\left(\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}\right)}_{\mathrm{ice}}= {\left(\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}\right)}_{\mathrm{water}} \\ \frac{K_{\mathrm{ice}}}{{\displaystyle \frac{x}{\mathrm{A}\times 10}}} = \frac{K_{\mathrm{water}}}{{\displaystyle \frac{1-x}{\mathrm{A}\times 4}}} \\ \frac{1.7\times 10}{x}=\frac{0.5\times 4}{1-x} \\ \frac{17}{x}=\frac{2}{1-x} \\ 17-17x=2x \\ \Rightarrow x=\frac{17}{19}=0.89 \mathrm{m} \\ \Rightarrow x=89 \mathrm{cm} \end{gathered}$$

Answer:

Thermal conductivity of rod AB, K_AB = 50 J/m-s-°C
Temperature of junction at A, T_A = 40°C
Thermal conductivity of rod BC, K_BC = 200 J/m-s-°C
Temperature of junction at B, T_B = 80°C
Thermal conductivity of rod BC, K_CA = 400 J/m-s-°C
Temperature of junction at C, T_C = 80°C
l = 20 cm = 20 × 10⁻² m
A = 1 cm² = 10⁻⁴ m²

(a) $\frac{{\left(\mathrm{\Delta }Q\right)}_{\mathrm{AB}}}{\mathrm{\Delta }t} = \frac{K_{\mathrm{AB}}A \left(T_{\mathrm{B}}-T_{\mathrm{A}}\right)}{l}$
      $$\begin{gathered} =\frac{50\times 1\times {10}^{-4} \left(40\right)}{20\times {10}^{-2}} \\ =1 \mathrm{W} \end{gathered}$$

(b) $\frac{{\left(\mathrm{\Delta }Q\right)}_{\mathrm{AC}}}{\mathrm{\Delta }t} = \frac{K_{\mathrm{AC}}\mathit{·}A\left(T_{\mathrm{C}}\mathit{-}{T}_{\mathrm{A}}\right)}{l}$
      $$\begin{gathered} =\frac{400\times 1\times {10}^{-4}\times \left(40\right)}{20\times {10}^{-2}} \\ =8 \mathrm{W} \end{gathered}$$

(c) $\frac{{\left(\mathrm{\Delta }Q\right)}_{\mathrm{BC}}}{\mathrm{\Delta }t} = \frac{\left(K_{\mathrm{BC}}\right)A \left(T_{\mathrm{B}}-T_{\mathrm{C}}\right)}{l}$
       = 0

Answer:

 

Let A be the area of cross section and K be the thermal conductivity of the material of the rod.
Let q₁ be the rate of flow of heat through a semicircular rod.
Rate of flow of heat is given by

$q_1 = \frac{dQ}{dt}=\frac{K\mathit{·}A \left(T_1-T_2\right)}{\mathrm{\pi } r}$

Let q₂ be the rate of flow of heat through a straight rod.
$q_{2 } = \frac{dQ}{dt}=\frac{KA\mathit{ }\left(T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}\right)}{2\mathit{ }r}$

Ratio of the rate of flow of heat through the 2 rods = $\frac{q_1}{q_2}=\frac{2 \mathrm{r}}{\mathrm{\pi } \mathrm{r}}=\frac{2}{\mathrm{\pi }}$

Answer:

Let the temperatures at the ends A and B be T_A and T_B, respectively.
Rate of flow of heat at end A of the rod is given by
$\frac{{\mathrm{dQ}}_{\mathrm{A}}}{\mathrm{d}t} = K\mathrm{A}.\frac{\mathrm{d}}{\mathrm{d}l}\left(T_{\mathrm{A}}\right)$
Rate of flow of heat at end B of the rod is given by
$\frac{d{Q}_{\mathrm{B}}}{\mathrm{d}t} = K\mathrm{A}.\frac{\mathrm{d}}{\mathrm{d}l}\left(T_{\mathrm{B}}\right)$
Heat absorbed by the rod = ms∆T
Here, s is the specific heat of the rod and ∆T is the temperature difference between ends A and B.

Rate of heat absorption by the rod is given by
$$\begin{gathered} \frac{\mathrm{d}Q}{\mathrm{d}t} = ms\frac{\mathrm{d}T}{\mathrm{d}t} \\ \therefore ms\frac{\mathrm{dT}}{\mathrm{d}t} = K\mathrm{A}.\frac{{\mathrm{dT}}_{\mathrm{A}}}{\mathrm{d}l}-K\mathrm{A}.\frac{{\mathrm{dT}}_{\mathrm{B}}}{\mathrm{d}l} \\ \Rightarrow \left(0.4\right).\frac{\mathrm{dT}}{\mathrm{d}t} = 200\times 1\times {10}^{-4} \left(5-2.5\right) \\ \frac{\mathrm{dT}}{\mathrm{d}t} = 12.5°\mathrm{C}/\sec \end{gathered}$$

Answer:

Inner radii = r = 1 cm = 10^–2 m
Outer radii = R = 1.2 cm = 1.2 × 10^–2 m
Length of the tube, l = 50 cm = 0.5 m
Thermal conductivity, k = 0.15 Js^–1 m^–1 °C^–1

Top View


Let us consider a cylindrical shell of length l, radius x and thickness dx.
Rate of flow of heat q$=\frac{d\mathrm{Q}}{dt}$
$\frac{dQ}{dt} = \frac{-KA∆T}{dx}$

Here, the negative sign indicates that the rate of heat flow decreases as x increases.
$$\begin{gathered} q=-K\left(2\mathrm{\pi }xl\right).\frac{\mathrm{dT}}{\mathrm{d}x} \\ {\int }_r^R \frac{\mathrm{d}x}{x}=-\frac{2\mathrm{\pi }Kl}{q}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}dT \\ {\left[\ln \left(x\right)\right]}_r^R = -\frac{2\mathrm{\pi K}l}{q}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \\ \Rightarrow q = \frac{2\mathrm{\pi }kl\left(T_1-T_2\right)}{\ln \left({\displaystyle \frac{R}{r}}\right)} \\ q = \frac{2\mathrm{\pi }\times 0.15\times 0.5\left(90\right)}{\ln \left({\displaystyle \frac{1.2\times {10}^{-2}}{1\times {10}^{-2}}}\right)} \\ q = 262.9 \mathrm{J}/\sec \end{gathered}$$

Answer:

Let $\frac{d\theta }{dt}$ be the rate of flow of heat.
Consider an annular ring of radius r and thickness dr.
Rate of flow of heat is given by
$\frac{\mathrm{d\theta }}{\mathrm{d}t} = -K\left(2\mathrm{\pi }rd\right).\frac{\mathrm{d\theta }}{\mathrm{d}r}$
Rate of flow of heat is constant.
∴ $\frac{\mathrm{d\theta }}{\mathrm{d}t} = i$ 
$$\begin{gathered} i = -K\left(2\mathrm{\pi }r.d\right)\frac{\mathrm{d\theta }}{\mathrm{d}r} \\ {\int }_{r_1}^{r_2} \frac{\mathrm{d}r}{r}=-\frac{2Kd}{i} {\int }_{{\theta }_1}^{{\theta }_2} d\mathrm{\theta } \\ {\left[\ln \left(r\right)\right]}_{r_1}^{r_2}=-\frac{2\mathrm{\pi }Kd}{i}\left[{\theta }_2-{\theta }_1\right] \\ i=\frac{2\mathrm{\pi }Kd \left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)}{\ln \left({\displaystyle \frac{r_2}{r_1}}\right)} \end{gathered}$$

Answer:

(a)  When the flat ends are maintained at temperatures T₁ and T₂ (where T₂ > T₁):

Area of cross section through which heat is flowing, $\mathrm{A}=\mathrm{\pi }\left({\mathrm{R}}_2^2-{\mathrm{R}}_1^2\right)$

​Rate of flow of heat$=\frac{d\theta }{dt}$
                         
                          $$\begin{gathered} =\frac{KA.∆T}{l} \\ =\frac{K\mathrm{\pi }\left({\mathrm{R}}_2^2-{\mathrm{R}}_1^2\right) \left({\mathrm{T}}_2-{\mathrm{T}}_1\right)}{l} \end{gathered}$$

(b)

When the inside of the tube is maintained at temperature T₁ and the outside is maintained at T_2:

Let us consider a cylindrical shell of radius r and thickness dr.
Rate of flow of heat, q$=KA.\frac{d\mathrm{T}}{dr}$
$$\begin{gathered} q = KA.\frac{\mathrm{dT}}{\mathrm{d}r} \\ q = K\left(2\mathrm{\pi }rl\right)\frac{\mathrm{dT}}{\mathrm{d}r} \\ {\int }_{{\mathrm{R}}_1}^{{\mathrm{R}}_2} \frac{dr}{r} = \frac{2\mathrm{\pi }Kl}{q}{\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2} d\mathrm{T} \\ {\left[\ln \left(r\right)\right]}_{{\mathrm{R}}_1}^{{\mathrm{R}}_2} = \frac{2\mathrm{\pi }Kl}{q} \left[{\mathrm{T}}_2-{\mathrm{T}}_1\right] \\ \ln \left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right) = \frac{2\mathrm{\pi }Kl}{q}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \\ q = \frac{2\mathrm{\pi }Kl \left({\mathrm{T}}_2-{\mathrm{T}}_1\right)}{\ln \left({\displaystyle \frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}}\right)} \end{gathered}$$

Answer:

It is equivalent to the series combination of 2 resistors.
∴ R_S = R₁ +R₂
Resistance of a conducting slab, $R=\frac{L}{KA}$
R_S = R₁ + R₂
$$\begin{gathered} \frac{L_1+L_2}{K_{\mathrm{s}}.A}=\frac{L_1}{K_{1}A}+\frac{L_2}{K_{2}A} \\ \frac{L_1+L_2}{K_{\mathrm{S}}}=\frac{L_1}{K_1}+\frac{L_2}{K_2} \\ \frac{L_1+L_2}{K_{\mathrm{S}}}=\frac{L_1{K}_2+L_2{K}_1}{K_1\times K_2} \\ {K}_{\mathrm{S}}=\frac{\left(L_1+L_2\right) \left(K_1 K_2\right)}{L_1 K_2+L_2{K}_1} \end{gathered}$$

Answer:

Rods are connected in series, so the rate of flow of heat is same.

q₁ = q₂

Rate of flow of heat is given by
$q = \frac{dQ}{dt} = \frac{\mathrm{Temperature} \mathrm{difference}}{\mathrm{Thermal} \mathrm{resistance}}$

As q₁ = q₂,
$$\begin{gathered} \Rightarrow \frac{T-0}{R_{\mathrm{cu}}} = \frac{100-T}{R_{\mathrm{S}}} \\ A{K}_1\left(\frac{T-0}{l}\right) = \left(\frac{100-T}{l}\right)K_{2}A \\ \Rightarrow 390\mathrm{T} = \left(100-\mathrm{T}\right)46 \\ \Rightarrow \mathrm{T} = 10.6°\mathrm{C} \end{gathered}$$

Answer:

q₁ and q₂ are heat currents. In other words, they are the rates of flow of heat through aluminium and copper rod, respectively.

Applying KVL at the hot junction, we get
q = q₁ + q₂
Rate of heat flow, q = $\frac{KA∆T}{l}$

As q = q₁ + q₂,

$$\begin{gathered} \frac{K_{\mathrm{P}}A\left(T_1-T_2\right)}{l}=\frac{K_{1}A\left(T_1-T_2\right)}{l}+\frac{K_{2}A\left(T_1-T_2\right)}{l} \\ Kp=K_1+K_2 \\ =390+200=590 \mathrm{W}/\mathrm{m}°\mathrm{C} \\ \Rightarrow q=\frac{K_{P}A\left(T_1-T_2\right)}{l} \\ q=\frac{590\times {10}^{-4} \left(60-20\right)}{1} \\ q=2.36 \mathrm{W} \end{gathered}$$

Answer:

Area of cross section, A = 0.20 cm² = 0.2 × 10^–4 m²
Thermal conductivity of aluminium, K_Al = 200 W/​m ​°C
Thermal conductivity of copper, K_Cu = 400 W/m​°C
Total heat flowing per second = q_Al + q_Cu
                                         $$\begin{gathered} =\frac{K_{\mathrm{Al}}\mathit{\times }A\mathit{\times }\left(80-40\right)}{l}+\frac{K_{\mathrm{Cu}}\times \mathrm{A}\times \left(80-40\right)}{l} \\ =\frac{200\times 0.2\times {10}^{-4}\times 40}{0.2}+\frac{400\times 0.2\times {10}^{-4}\times 40}{0.2} \\ =8\times {10}^{-1}+16\times {10}^{-1} \\ =24\times {10}^{-1} \\ =2.4 \mathrm{J}/\sec \end{gathered}$$
Heat drawn in 1 minute = 2.4 × 60 = 144 J

Answer:

$$\begin{gathered} R_{\mathrm{BC}}=\frac{l}{KA}=\frac{5}{KA} \\ {R}_{\mathrm{CD}}=\frac{l}{KA}=\frac{60}{KA} \\ {R}_{\mathrm{DE}}=\frac{5}{KA}, R_{\mathrm{AB}}=\frac{20}{KA}, R_{\mathrm{EF}}=\frac{20}{KA} \\ \mathrm{Let}: \\ R_1=R_{\mathrm{BC}}+R_{\mathrm{CD}}+R_{\mathrm{DE}}=\frac{70}{KA} \\ \mathrm{Let}: \\ R_{\mathrm{BE}}=\frac{60}{KA}=R_2 \end{gathered}$$

q = q₁ + q₂              ...(1)

R₁ and R₂ are in parallel, so total heat across R₁ and R₂ will be same.

⇒ q₁R₁ = q₂R₂
$$\begin{gathered} q_1\times \frac{70}{KA} = q_2\times \frac{60}{KA} \\ 7q_1 = 6q_2 \\ \frac{7q_1}{6} = q_2 ...\left(2\right) \end{gathered}$$
From equation (1) and (2),
$$\begin{gathered} q = q_1+\frac{7q_1}{6} \\ q = \frac{13q_1}{6} \\ q = 130 \mathrm{J} \\ 130 = \frac{13q_1}{6} \\ {q}_1 = 60 \mathrm{J}/\sec \end{gathered}$$

Answer:

Resistance of any branch, R = $\frac{l}{KA}$

Here, K is the thermal conductivity, A is the area of cross section and l is the length of the conductor.
$$\begin{gathered} R_{\mathrm{BC}}=\frac{l}{780.\mathrm{A}}=\frac{5\times {10}^{-2}}{780.\mathrm{A}} \\ {R}_{\mathrm{CD}}=\frac{60\times {10}^{-2}}{780.\mathrm{A}} \\ {R}_{\mathrm{DE}}=\frac{5\times {10}^{-2}}{780.\mathrm{A}} \\ {R}_{\mathrm{AB}}=\frac{20\times {10}^{-2}}{390.\mathrm{A}} \\ {R}_{\mathrm{EF}}=\frac{20\times {10}^{-2}}{390.\mathrm{A}} \\ {R}_{\mathrm{BE}}=\frac{60\times {10}^{-2}}{390\times \mathrm{A}} \end{gathered}$$

$R_1=R_{\mathrm{BC}}+R_{\mathrm{CD}}+R_{\mathrm{DE}}=\frac{70\times {10}^{-2}}{780\times \mathrm{A}}$

$R_{BE}=R_2=\frac{60\times {10}^{-2}}{390\times A}$
Since R₁ and R₂ are in parallel, the amount of heat flowing through them will be same.
$$\begin{gathered} q_1{R}_1=q_2{R}_2 \\ \frac{q_1}{q_2}=\frac{R_2}{R_1} \\ =\frac{60\times {10}^{-2}\times 780\times \mathrm{A}}{390\times \mathrm{A}\times 70\times {10}^{-2}} \\ =\frac{12}{7} \\ \Rightarrow \frac{q_1}{q_2}=\frac{12}{7} \end{gathered}$$

Answer:

(a)

Length, l = 2 mm = 0.0002 m

$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat} = \frac{∆T}{l/KA} \\ =\frac{KA.∆T}{l} \\ =\frac{1\times 2\times \left(40-32\right)}{2\times {10}^{-3}} \\ = 8000 \mathrm{J}/\sec \end{gathered}$$

(b)

Resistance of glass, $R_{\mathrm{g}} = \frac{l}{K_{\mathrm{g}}.A}$
Resistance of air, $R_{\mathrm{A}} = \frac{l}{K_{\mathrm{A}}.A}$

From the circuit diagram, we can find that all the resistors are connected in series.
                              
$$\begin{gathered} R_s=R_g+R_A+R_g \\ =\frac{{10}^{-3}}{2}\left(\frac{2}{K_{\mathrm{g}}}+\frac{1}{K_{\mathrm{A}}}\right) \\ =\frac{{10}^{-3}}{2}\left(\frac{2}{1}+\frac{1}{0.025}\right) \\ =\frac{{10}^{-3}}{2}\times \frac{\left(2\times 0.025+1\right)}{0.025} \end{gathered}$$
$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}, q= \frac{∆T}{R_s} \\ = \frac{T_1-T_2}{R_s} \\ =\frac{\left(40-32\right)\times 2\times 0.025}{{40}^{-3}\times \left(2\times 0.025+1\right)} \\ = 381W \end{gathered}$$

Answer:

As the rods are connected in series, the rate of flow of heat will be same in both the cases.

Case 1:
Rate of flow of heat is given by
$\frac{dQ}{dt} = \frac{KA∆T}{l}$
Rate of heat flow in rod P will be same as that in rod Q.

$$\begin{gathered} \therefore \frac{K_{\mathrm{p}}\mathit{\times }A\mathit{\times }\left(100-70\right)}{l} = \frac{K_{\mathrm{Q}}\times A\times \left(70-0\right)}{l} \\ \Rightarrow 30 K_{\mathrm{P}}\mathit{ }= 70 K_{\mathrm{Q}} \\ \Rightarrow K_{\mathrm{Q}} = \frac{3}{7} K_{\mathrm{P}} ......\left(\mathrm{i}\right) \end{gathered}$$
              
Case 2:
Again, the rate of flow of heat will be same in rod P and Q.

$$\begin{gathered} \therefore \frac{K_{\mathrm{Q}}\times A\times \left(100-\mathrm{T}\right)}{l} = \frac{K_{\mathrm{P}}\times A\times \left(\mathrm{T}-0\right)}{l} \\ 100 K_{\mathrm{Q}}-K_{\mathrm{Q}}\mathrm{T}=K_{\mathrm{P}}\mathrm{T} \\ 100 K_{\mathrm{Q}}-K_{\mathrm{Q}}T=\frac{70}{30}{K}_{\mathrm{Q}}T [\mathrm{using} (\mathrm{i}\left)\right] \\ 100-T=\frac{7}{3}T \\ 100 = \frac{10}{3}T \\ \Rightarrow T = 30°\mathrm{C} \end{gathered}$$
               

Answer:

For arrangement (a),


Temperature of the hot end ,T₁ = 100°C

Temperature of the cold end ,T₂ = 0°C
R_s = R₁ + R₂ + R₃
$=\frac{l}{K_{\mathrm{Al}}\mathrm{A}}+\frac{l}{K_{\mathrm{cu}}\mathrm{A}}+\frac{l}{K_{\mathit{Al}}A}$

$$\begin{gathered} =\frac{\mathit{l}}{\mathit{A}}\left(\frac{1}{200}+\frac{1}{400}+\frac{1}{200}\right) \\ =\frac{l}{\mathrm{A}}\left(\frac{5}{400}\right) \\ =\frac{l}{\mathrm{A}}\times \frac{1}{80} \end{gathered}$$
$$\begin{gathered} \frac{\mathrm{d}Q}{dt} =\mathit{ }q = \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat} = \frac{T_1-T_2}{R_{\mathrm{S}}} \\ =\frac{100-0}{{\displaystyle \frac{l}{\mathrm{A}}\times \frac{1}{80}}} \\ \mathrm{Given}: \\ q=40 \mathrm{W} \\ 40=\frac{100}{{\displaystyle \frac{l}{\mathrm{A}}}\times {\displaystyle \frac{1}{80}}} \\ \Rightarrow \frac{l}{\mathrm{A}}=200 \\ \Rightarrow \frac{\mathrm{A}}{l}=\frac{1}{200} \end{gathered}$$

For arrangement (b),


  $$\begin{gathered} R_{\mathrm{net}}\mathit{ }= R_{\mathrm{Al}}+\frac{R_{\mathrm{Cu}}\times R_{\mathrm{Al}}}{{\mathrm{R}}_{\mathrm{Cu}}+{\mathrm{R}}_{\mathrm{Al}}} \\ =\frac{l}{K_{\mathrm{Al}}A}+\frac{{\displaystyle \frac{l}{K_{\mathrm{Cu}}A}}\times {\displaystyle \frac{l}{K_{\mathrm{Al}}l}}}{{\displaystyle \frac{l}{K_{\mathrm{Cu}}A}}+{\displaystyle \frac{l}{K_{\mathit{Al}}\mathrm{A}}}} \\ =\frac{\mathit{l}}{\mathit{A}\mathit{·}{\mathit{K}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{l}}{\mathit{A}\mathit{ }\left({\mathit{K}}_{\mathit{Al}}\mathit{+}{\mathit{K}}_{\mathit{Cu}}\right)} \\ =\frac{l}{\mathrm{A}}\left(\frac{1}{200}+\frac{1}{200+400}\right) \\ =\frac{l}{\mathrm{A}}\left(\frac{1}{200}+\frac{1}{600}\right) \\ =\frac{4}{600}.\frac{l}{\mathrm{A}} \end{gathered}$$

Rate of flow of heat is given by
$$\begin{gathered} q=\frac{{\mathrm{T}}_1-{\mathrm{T}}_2}{{\mathrm{R}}_{\mathrm{net}}} \\ =\frac{\left(100-0\right)}{4 l}600 \mathrm{A} \\ =\frac{100\times 600}{4}\times \frac{1}{200} \\ =75 \mathrm{W} \end{gathered}$$

For arrangement (c),

 

 $$\begin{gathered} \frac{\mathit{1}}{{\mathit{R}}_{\mathit{net}}} \mathit{=}\mathit{ }\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Cu}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}} \\ =\frac{K_{\mathrm{Al}}A}{l}+\frac{K_{\mathrm{Cu}}A}{l}+\frac{K_{\mathrm{Al}}A}{l} \\ \frac{1}{R_{\mathrm{net}}} = \left(K_{\mathrm{Al}}+K_{\mathrm{Cu}}+K_{\mathrm{Al}}\right) \frac{\mathit{A}}{\mathit{l}} \\ \frac{1}{R_{\mathrm{net}}}=\left(200+400+200\right)\times \frac{1}{200} \\ \Rightarrow R_{\mathrm{net}}\mathit{ }= \frac{200}{800} \\ = \frac{1}{4} \end{gathered}$$

$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{heat} \mathrm{flow}=\frac{\Delta T}{R_{\mathrm{net}}} \\ = \frac{100}{{\displaystyle \frac{1}{4}}} \\ =400 \mathrm{W} \end{gathered}$$

Answer:

 
Let the temperature at junction B be T.
Let q₁, q₂ and q₃ be the heat currents, i.e. rate of flow of heat per unit time in AB, BCE and BDF, respectively.
From the diagram, we can see that
q₁ = q₂ + q₃
The rate of flow of heat is givem by
$q = \frac{KA∆T}{l}$
Using this tn the above equation, we get

$$\begin{gathered} \frac{KA \left(T_1-T\right)}{l} = \frac{KA \left(T-T_3\right)}{{\displaystyle \frac{3l}{2}}}+\frac{KA \left(T-T_2\right)}{{\displaystyle \frac{3l}{2}}} \\ \Rightarrow T_{\mathit{1}}\mathit{-}T\mathit{ }\mathit{=}\mathit{ }\frac{\mathit{2}\mathit{ }\left(\mathit{T}\mathit{-}{\mathit{T}}_{\mathit{3}}\right)}{\mathit{3}}\mathit{+}\frac{\mathit{2}\mathit{ }\left(\mathit{T}\mathit{-}{\mathit{T}}_{\mathit{2}}\right)}{\mathit{3}} \\ \mathit{\Rightarrow }3\left({\mathit{T}}_{\mathit{1}}\mathit{-}\mathit{T}\right) \mathit{=}\mathit{ }2T\mathit{-}2T_{\mathit{3}}\mathit{+}2T\mathit{-}2T_{\mathit{2}} \\ \mathit{\Rightarrow }T\mathit{ }\mathit{=}\mathit{ }\frac{\mathit{-}\mathit{3}{\mathit{T}}_{\mathit{1}}\mathit{+}\mathit{2}\mathit{ }\left({\mathit{T}}_{\mathit{2}}\mathit{+}{\mathit{T}}_{\mathit{3}}\right)}{\mathit{7}} \end{gathered}$$

Answer:

Given:
K_A = K_C = K₀
K_B = K_D = 2K₀
K_E = 3K₀, K_F = 4K₀
K₉= 5K₀
Here, K denotes the thermal conductivity of the respective rods.

In steady state, temperature at the ends of rod F will be same.

Rate of flow of heat through rod A + rod C = Rate of flow of heat through rod B + rod D

$$\begin{gathered} \frac{K_{\mathrm{A}}·A·\left(T-T_1\right)}{l}+\frac{K_{\mathrm{C}}·A \left(T-T_1\right)}{l} = \frac{K_{\mathrm{B}}·A·\left(T_2-T\right)}{l}+\frac{K_{\mathrm{D}}·A \left(T_2-T\right)}{l} \\ 2K_{0 }\left(T-T_1\right)\mathit{ }= 2\times 2 K_{0 }\left(T_2-T\right) \\ \mathrm{Temp} \mathrm{of} \mathrm{rod} \mathrm{F} = T\mathit{ }= \frac{T_1+2T_2}{3} \end{gathered}$$
(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T₂ is given by
Rate of flow of heat, q = $\frac{∆T}{\mathrm{Thermal} \mathrm{resistance}}$
First, we will find the effective thermal resistance of the circuit.
From the diagram, we can see that it forms a balanced Wheatstone bridge.
Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.
Hence, for simplification, we can remove this branch.

From the diagram, we find that R_A and R_B are connected in series.
∴ R_AB = R_A + R_B
R_C and R_D are also connected in series.
∴ R_CD = R_C + RD
Then, R_AB and R_CD are in parallel connection.
$$\begin{gathered} R_{\mathrm{A}} = \frac{l}{K_{0}A} \\ {R}_{\mathrm{B}} = \frac{l}{2K_{0}A} \\ {R}_{\mathrm{C}} = \frac{l}{K_{0}A} \\ {R}_{\mathrm{D}} = \frac{l}{2K_{0}A} \\ {R}_{\mathrm{AB} }= \frac{3l}{2K_{0}A} \\ {R}_{\mathrm{CD}} = \frac{3l}{2K_{0}A} \\ {R}_{\mathrm{eff}} = \frac{\frac{3l}{2K_{0}A}\times \frac{3l}{2K_{0}A}}{\frac{3l}{2K_{0}A}+\frac{3l}{2K_{0}A}} \\ = \frac{3l}{4K_{0}A} \end{gathered}$$
$$\begin{gathered} \Rightarrow q = \frac{∆T}{R_{eff}} \\ =\frac{T_1-T_2}{{\displaystyle \frac{3l}{4K_{0}A}}} \\ =\frac{4K_{0}A(T_1-T_2)}{3l} \end{gathered}$$

Answer:

We can infer from the diagram that ΔPQR is similar to ΔPST.
So, by the property of similar triangles,
$$\begin{gathered} \frac{x}{l} = \frac{r-r_1}{r_2-r_1} \\ \Rightarrow \left(r_2-r_1\right) \frac{x}{l} = r-r_1 \\ \Rightarrow r = r_1+\left(r_2-r_1\right) \frac{x}{l} \\ \mathrm{Let}: \\ \frac{r_2-r_1}{l} = a \\ \Rightarrow r = ax+r_1 ..........\left(\mathrm{i}\right) \end{gathered}$$

Thermal resistance is given by
$$\begin{gathered} \mathrm{dR} = \frac{dx}{\mathrm{K}·\mathrm{A}} \\ \Rightarrow \mathrm{dR}=\frac{dx}{\mathrm{K}·\mathrm{\pi }{r}^2} \\ \mathrm{dR}=\frac{dx}{\mathrm{K}·\mathrm{\pi } {\left(ax+r_1\right)}^2} \\ \underset{0}{\overset{\mathrm{R}}{\int }}\mathrm{dR}=\frac{1}{\mathrm{K\pi }}\underset{0}{\overset{l}{\int }}\frac{dx}{{\left(ax+r_1\right)}^2} \\ \mathrm{R}=-\frac{1}{a \mathrm{K\pi }}{\left[\frac{1}{ax+r_1}\right]}_0^l \\ \mathrm{R}=\frac{-1}{a \mathrm{K\pi }}\left[\frac{1}{al+r_1}-\frac{1}{r_1}\right] \\ \mathrm{R}=-\frac{1}{a \mathrm{K\pi }}\left[\frac{1}{\left({\displaystyle \frac{r_2-r_1}{l}}\right) l+r_1}-\frac{1}{r_1}\right] \\ \mathrm{R}=-\frac{1}{a \mathrm{K\pi }}\left[\frac{1}{r_2}-\frac{1}{r_1}\right] \\ \mathrm{R}=\frac{-1}{\mathrm{K\pi }\left({\displaystyle \frac{r_2-r_1}{l}}\right)}\times \left(\frac{r_1-r_2}{r_1 r_2}\right) \\ \mathrm{R}=\frac{l}{\mathrm{K\pi }{r}_1{r}_2} \\ \mathrm{Rate} \mathrm{of} \mathrm{flow} \mathrm{of} \mathrm{heat}=\frac{\mathrm{\Delta } \mathrm{Q}}{\mathrm{R}} \\ \Rightarrow q = \frac{\left({\mathrm{Q}}_2-{\mathrm{Q}}_1\right)}{l}\mathrm{K\pi } r_1{r}_2 \end{gathered}$$