Mix Examples-Heat Transfer Questions in English

(D) In the variable state of heat conduction,the rate of heat flow is governed by the thermal diffusivity of the material. The thermal diffusivity $\alpha$ is defined as $\alpha = \frac{K}{\rho c}$,where $K$ is the thermal conductivity,$\rho$ is the density,and $c$ is the specific heat capacity.
Since the rate of change of temperature in the variable state depends on the thermal diffusivity,the rate of heat flow is influenced by all three factors: thermal conductivity $(K)$,density $(\rho)$,and specific heat $(c)$.
Therefore,the correct option is $D$.

(D) In a closed room,the air inside acts as a fluid. When a heat source is present,the air near the source gets heated,becomes less dense,and rises,while cooler,denser air moves in to take its place. This continuous circulation of air,known as convection,is the primary mechanism for heat transfer throughout the room. While conduction occurs through walls and radiation occurs from surfaces,the overall distribution of heat in the room is dominated by convection. Therefore,the correct answer is $D$.

(B) The correct answer is $B$.
$1$. At high altitudes,the atmosphere is thinner,which reduces the greenhouse effect and heat retention compared to the foot of the mountain.
$2$. The intensity of solar radiation reaching the surface depends on the angle of incidence. At the top of a mountain,the air temperature is significantly lower due to the adiabatic lapse rate.
$3$. Furthermore,the object at the top is exposed to higher wind speeds and lower ambient temperatures,leading to greater convective heat loss compared to the object at the foot.
$4$. Therefore,the object at the top of the mountain will register a lower temperature than the identical object placed at the foot.

(A) The rate of heat flow through a rod is given by the formula: $\frac{dQ}{dt} = \frac{KA \Delta \theta}{l}$.
Here,$K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta \theta$ is the temperature difference,and $l$ is the length of the rod.
Since both rods are made of the same material,$K$ is the same for both.
They have the same cross-sectional area $A$,and the points $A$ and $B$ are maintained at the same temperature difference $\Delta \theta$ for both paths.
Therefore,the rate of heat flow is inversely proportional to the length of the rod: $\frac{dQ}{dt} \propto \frac{1}{l}$.
The length of the straight rod is $l_{straight} = 2r$,where $r$ is the radius.
The length of the semi-circular rod is $l_{semi} = \pi r$.
Thus,the ratio of heat transferred is:
$\frac{(dQ/dt)_{semi}}{(dQ/dt)_{straight}} = \frac{l_{straight}}{l_{semi}} = \frac{2r}{\pi r} = \frac{2}{\pi}$.

(C) Since the heat flows along the length of the cylinders,the two cylinders are in parallel configuration.
The equivalent thermal conductivity $K_{eq}$ for parallel combination is given by $K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$.
Here,$A_1$ is the cross-sectional area of the inner cylinder: $A_1 = \pi R^2$.
$A_2$ is the cross-sectional area of the outer cylindrical shell: $A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$.
The total area $A = A_1 + A_2 = \pi R^2 + 3\pi R^2 = 4\pi R^2$.
Substituting these values into the formula:
$K_{eq} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{4\pi R^2} = \frac{K_1 + 3K_2}{4}$.

(A) The heat required to melt $500 \ g$ of ice at $0^\circ C$ is given by $Q = mL$,where $m = 500 \ g$ and $L = 80 \ cal/g$.
$Q = 500 \times 80 = 40,000 \ cal$.
The rate of heat flow through the cork is given by $\frac{Q}{t} = \frac{KA\Delta\theta}{\Delta x}$,where $K = 0.0075 \ cal/(cm \cdot s \cdot ^\circ C)$,$A = 75 \ cm^2$,$\Delta\theta = (40 - 0) = 40^\circ C$,and $\Delta x = 5 \ cm$.
Substituting the values,we get:
$\frac{40,000}{t} = \frac{0.0075 \times 75 \times 40}{5}$
$\frac{40,000}{t} = 0.0075 \times 15 \times 40$
$\frac{40,000}{t} = 4.5$
$t = \frac{40,000}{4.5} \approx 8888.89 \ s$.
To convert the time into hours,divide by $3600$:
$t = \frac{8888.89}{3600} \approx 2.47 \ hr$.

(B) Let the temperature of the junction be $\theta$. According to the principle of conservation of energy,the heat flowing into the junction must equal the heat flowing out of the junction.
$H = H_1 + H_2$
Using the formula for heat current $H = \frac{KA(\Delta T)}{l}$,where $K$ is thermal conductivity,$A$ is area,and $l$ is length:
$\frac{3K \cdot A \cdot (100 - \theta)}{l} = \frac{2K \cdot A \cdot (\theta - 50)}{l} + \frac{K \cdot A \cdot (\theta - 20)}{l}$
Since $A$ and $l$ are the same for all rods,they cancel out:
$3(100 - \theta) = 2(\theta - 50) + 1(\theta - 20)$
$300 - 3\theta = 2\theta - 100 + \theta - 20$
$300 - 3\theta = 3\theta - 120$
$420 = 6\theta$
$\theta = 70^oC$
Thus,the temperature of the junction is $70^oC$.

(D) According to Stefan-Boltzmann law,the total radiant power $P = e A \sigma T^4$. Since $P_A = P_B$ and $A_A = A_B$,we have $e_A T_A^4 = e_B T_B^4$.
Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802\;K$.
$T_B = T_A \left( \frac{e_A}{e_B} \right)^{1/4} = 5802 \times \left( \frac{0.01}{0.81} \right)^{1/4} = 5802 \times \left( \frac{1}{81} \right)^{1/4} = 5802 \times \frac{1}{3} = 1934\;K$.
From Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (constant).
$\lambda_A = \frac{b}{T_A}$ and $\lambda_B = \frac{b}{T_B}$.
Given $\lambda_B - \lambda_A = 1.00\;\mu m$.
$\frac{b}{T_B} - \frac{b}{T_A} = 1.00\;\mu m$.
$b \left( \frac{1}{1934} - \frac{1}{5802} \right) = 1.00\;\mu m$.
$b \left( \frac{3 - 1}{5802} \right) = 1.00\;\mu m \Rightarrow b = \frac{5802}{2} = 2901\;\mu m \cdot K$.
Now,$\lambda_B = \frac{b}{T_B} = \frac{2901}{1934} = 1.5\;\mu m$.
Thus,both $(a)$ and $(b)$ are correct.

(B) The power radiated by a black body is given by the Stefan-Boltzmann law: $P = A \sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
Since the discs are coated with carbon black,they act as black bodies $(\varepsilon = 1)$.
The area of a disc is $A = \pi r^2$,so $P \propto r^2 T^4$.
According to Wien's displacement law,$\lambda_m T = b$ (constant),which implies $T \propto \frac{1}{\lambda_m}$.
Substituting this into the power equation: $P \propto r^2 \left(\frac{1}{\lambda_m}\right)^4 = \frac{r^2}{\lambda_m^4}$.
Now,calculating the relative values for $Q_A, Q_B,$ and $Q_C$:
$Q_A \propto \frac{2^2}{300^4} = \frac{4}{81 \times 10^8} \approx 0.049 \times 10^{-8}$
$Q_B \propto \frac{4^2}{400^4} = \frac{16}{256 \times 10^8} = 0.0625 \times 10^{-8}$
$Q_C \propto \frac{6^2}{500^4} = \frac{36}{625 \times 10^8} \approx 0.0576 \times 10^{-8}$
Comparing these values,$Q_B$ is the maximum.

(C) Let the thermal resistance of each rod be $R$. The square consists of four rods $AC, CB, BD, DA$. The points $A$ and $B$ are maintained at temperatures $T_A = \sqrt{2}T$ and $T_B = T$.
Due to the symmetry of the circuit,the heat current $H$ flowing from $A$ splits equally into two paths: $AC-CB$ and $AD-DB$.
Let $T_C$ and $T_D$ be the temperatures at points $C$ and $D$ respectively.
For path $AC-CB$: The heat current is $H_1 = \frac{T_A - T_C}{R} = \frac{T_C - T_B}{R}$. This implies $T_A - T_C = T_C - T_B$,so $T_C = \frac{T_A + T_B}{2}$.
For path $AD-DB$: The heat current is $H_2 = \frac{T_A - T_D}{R} = \frac{T_D - T_B}{R}$. This implies $T_A - T_D = T_D - T_B$,so $T_D = \frac{T_A + T_B}{2}$.
Since $T_C = T_D$,the temperature difference between $C$ and $D$ is $T_C - T_D = 0$.

(A) Given: $\lambda_m = 0.48 \ \mu m = 0.48 \times 10^{-6} \ m$,$r = 6.96 \times 10^8 \ m$,$\sigma = 5.67 \times 10^{-8} \ W/m^2K^4$,$b = 0.293 \times 10^{-2} \ m \cdot K$.
$1$. Calculate the temperature of the sun using Wien's displacement law: $T = \frac{b}{\lambda_m} = \frac{0.293 \times 10^{-2}}{0.48 \times 10^{-6}} \approx 6104 \ K$.
$2$. Calculate the surface area of the sun: $A = 4\pi r^2 = 4 \times 3.1416 \times (6.96 \times 10^8)^2 \approx 6.084 \times 10^{18} \ m^2$.
$3$. Calculate the power (energy per second) radiated by the sun using Stefan-Boltzmann law: $P = A \sigma T^4 = (6.084 \times 10^{18}) \times (5.67 \times 10^{-8}) \times (6104)^4 \approx 4.789 \times 10^{26} \ J/s$.
$4$. Calculate the mass loss per second using Einstein's mass-energy equivalence: $\Delta m = \frac{P}{c^2} = \frac{4.789 \times 10^{26}}{(3 \times 10^8)^2} \approx 5.32 \times 10^9 \ kg/s$.

(B) The formula for heat conduction is $Q = K A \frac{\Delta \theta}{l} t$.
From this,we can see that for a constant amount of heat $Q$ and temperature difference $\Delta \theta$,the time $t$ is proportional to $\frac{l}{A}$,i.e.,$t \propto \frac{l}{A}$.
In Figure $1$,two rods of length $l$ and area $A$ are connected in series. The total length is $2l$ and the area is $A$. So,$t_1 \propto \frac{2l}{A}$.
In Figure $2$,the two rods are connected in parallel. The length is $l$ and the total area is $2A$. So,$t_2 \propto \frac{l}{2A}$.
Taking the ratio:
$\frac{t_1}{t_2} = \frac{2l/A}{l/2A} = \frac{2l}{A} \times \frac{2A}{l} = 4$.
Given $t_1 = 12 \, sec$,we have $t_2 = \frac{t_1}{4} = \frac{12}{4} = 3 \, sec$.

(C) In steady state, the rate of heat flow $H = -kA(dT/dx)$ is constant throughout the rod.
Given $k(x) = 2K - (K/L)x = K(2 - x/L)$.
Thus, $H = -K(2 - x/L)A(dT/dx) = \text{constant} = C$.
Rearranging, $dT = -[C / (KA(2 - x/L))] dx$.
Integrating both sides, $T(x) = -[C / (KA)] \int [1 / (2 - x/L)] dx$.
$T(x) = [CL / (KA)] \ln(2 - x/L) + C_0$.
Since $T(x)$ involves a logarithmic function, the slope $dT/dx = -C / [KA(2 - x/L)]$ is not constant.
As $x$ increases from $0$ to $L$, the term $(2 - x/L)$ decreases from $2$ to $1$.
Therefore, the magnitude of the slope $|dT/dx| = C / [KA(2 - x/L)]$ increases as $x$ increases.
$A$ curve where the magnitude of the slope increases as $x$ increases is concave up. Thus, the graph is concave up.

(D) In a steady-state condition,the rate of heat flow $H = \frac{kA \Delta T}{L}$ is constant through all layers. Since the area $A$ and thickness $L$ are the same for all three layers,the temperature drop $\Delta T$ across each layer is inversely proportional to its thermal conductivity $k$ $(\Delta T = \frac{HL}{kA} \propto \frac{1}{k})$.
Since the thermal conductivity of brick $(k_b)$ is much greater than that of air $(k_a)$,the temperature drop across the air layer will be much larger than the temperature drop across each brick layer.
Therefore,the graph must show a steep linear decrease in temperature across the air layer and a gradual linear decrease across the brick layers. Option $(D)$ correctly represents this behavior.

(C) For an unlagged metal bar,heat is lost to the surroundings from the surface of the bar due to convection and radiation.
Let $T$ be the temperature at a distance $x$ from the end maintained at $T_1$. The steady-state heat conduction equation for an unlagged rod is given by $\frac{d^2T}{dx^2} = k(T - T_s)$,where $T_s$ is the surrounding temperature and $k$ is a constant related to the material and surface properties.
The solution to this differential equation is of the form $T(x) = T_s + A e^{mx} + B e^{-mx}$.
This results in a temperature profile that is concave upwards (exponential decay towards the ambient temperature).
Since the bar is losing heat to the surroundings along its length,the temperature gradient decreases as we move away from the hot end,resulting in a concave curve.
Therefore,the correct representation is a concave curve.

(A) Both the rods $AB$ and $CD$ are in a steady state.
This means the temperature gradient is constant and the temperature drops linearly with distance.
Temperature at the midpoint of $AB$ is $T_{P} = \frac{T_{A} + T_{B}}{2} = \frac{0^{\circ} C + 100^{\circ} C}{2} = 50^{\circ} C$.
Temperature at the midpoint of $CD$ is $T_{Q} = \frac{T_{C} + T_{D}}{2} = \frac{30^{\circ} C + 60^{\circ} C}{2} = 45^{\circ} C$.
Since the temperature at $P$ $(50^{\circ} C)$ is higher than the temperature at $Q$ $(45^{\circ} C)$,heat flows from $P$ to $Q$.

(D) According to Wien's displacement law,$\lambda_{\max} T = \text{constant}$.
Let the initial temperature be $T$ and the final temperature be $T'$.
Given $\lambda_{\max, 1} = \lambda_0$ and $\lambda_{\max, 2} = \frac{3}{4}\lambda_0$.
Using $\lambda_{\max, 1} T_1 = \lambda_{\max, 2} T_2$,we get $\lambda_0 T = \frac{3}{4}\lambda_0 T'$,which implies $T' = \frac{4}{3}T$.
According to the Stefan-Boltzmann law,the power radiated $P \propto T^4$.
Therefore,the ratio of the new power $P'$ to the initial power $P$ is:
$\frac{P'}{P} = \left(\frac{T'}{T}\right)^4 = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$.

(A) In steady-state heat conduction through a uniform metallic bar,the rate of heat flow $\frac{dQ}{dt}$ is constant throughout the bar.
According to Fourier's law of heat conduction:
$\frac{dQ}{dt} = -kA \frac{d\theta}{dx}$
Where $k$ is the thermal conductivity,$A$ is the cross-sectional area,and $\frac{d\theta}{dx}$ is the temperature gradient.
Since $\frac{dQ}{dt}$,$k$,and $A$ are constants,the temperature gradient $\frac{d\theta}{dx} = -\frac{1}{kA} \frac{dQ}{dt}$ must also be a constant.
This implies that the temperature $\theta$ decreases linearly with distance $x$ from the hot end.
Therefore,the graph of $\theta$ versus $x$ is a straight line with a negative slope,which corresponds to figure $A$.

(B) The rate of heat flow $Q$ is given by the formula:
$Q = \frac{KA(\theta_1 - \theta_2)}{l}$
where $K$ is the coefficient of thermal conductivity,$l$ is the length of the rod,and $A$ is the area of cross-section.
Let $T$ be the temperature at the junction where the three rods meet.
According to the principle of conservation of energy,the heat flowing into the junction through the copper rod must equal the heat flowing out through the brass and steel rods:
$Q_{\text{copper}} = Q_{\text{brass}} + Q_{\text{steel}}$
Substituting the given values:
$\frac{0.92 \times 4 \times (100 - T)}{46} = \frac{0.26 \times 4 \times (T - 0)}{13} + \frac{0.12 \times 4 \times (T - 0)}{12}$
Simplifying the equation:
$0.08 \times (100 - T) = 0.08 \times T + 0.04 \times T$
$8 - 0.08T = 0.12T$
$8 = 0.2T$
$T = \frac{8}{0.2} = 40^\circ C$
Now,calculate the rate of heat flow through the copper rod:
$Q_{\text{copper}} = \frac{0.92 \times 4 \times (100 - 40)}{46} = \frac{0.92 \times 4 \times 60}{46} = 0.02 \times 4 \times 60 = 4.8 \ cal \ s^{-1}$.

(D) According to Wien's displacement law: $\lambda_{m} \cdot T = b$,which implies $T \propto \frac{1}{\lambda_{m}}$.
Given $\lambda_{m1} = 20,000 \ \mathring{A}$ and $\lambda_{m2} = 10,000 \ \mathring{A}$.
Since $\lambda_{m}$ is halved,the temperature $T$ must double,i.e.,$T_{2} = 2T_{1}$.
According to the Stefan-Boltzmann law,the energy radiated per unit area per unit time is $e = \sigma T^{4}$.
Therefore,$\frac{e_{2}}{e_{1}} = \left(\frac{T_{2}}{T_{1}}\right)^{4} = (2)^{4} = 16$.
Given $e_{1} = 16 \ J \ m^{-2} \ s^{-1}$.
Thus,$e_{2} = 16 \times 16 = 256 \ J \ m^{-2} \ s^{-1}$.

(B) According to Wien's displacement law,the frequency $v_{max}$ at which the maximum intensity occurs is directly proportional to the absolute temperature $T$,i.e.,$v_{max} \propto T$.
Since the temperature increases from $T$ to $3T$,the new frequency $v'$ will be $v' = 3v$. Thus,statement $(ii)$ is correct.
According to the Stefan-Boltzmann law,the total energy $E$ radiated per unit time by a body is proportional to the fourth power of its absolute temperature,i.e.,$E \propto T^4$.
Given the initial energy is $E$ at temperature $T$,the new energy $E'$ at temperature $3T$ is $E' = E \times (3T/T)^4 = E \times 3^4 = 81E$. Thus,statement $(iii)$ is correct.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(D) $(i)$ According to Kirchhoff's law of radiation,good absorbers are good emitters. Since a body with high reflectivity is a poor absorber,it is also a poor emitter. This statement is $CORRECT$.
$(ii)$ Brass is a good conductor of heat,while wood is a poor conductor. On a chilly day,the brass tumbler conducts heat away from the hand much faster than the wooden tray,making it feel colder. This statement is $CORRECT$.
$(iii)$ The atmosphere traps heat via the greenhouse effect. Without it,the Earth would lose most of its heat through radiation,becoming inhospitably cold. This statement is $CORRECT$.
$(iv)$ Steam carries latent heat of vaporization,which is released upon condensation. This provides significantly more heat per unit mass compared to hot water,which only releases sensible heat. Thus,steam-based systems are more efficient. This statement is $CORRECT$.
Therefore,all statements are correct.

(A) According to Wien's displacement law,$\lambda_{\max} T = b$,where $b$ is Wien's constant.
Since $\lambda_{\max} = \frac{c}{f_{\max}}$,we have $\frac{c}{f_{\max}} T = b$,which implies $f_{\max} = \frac{c}{b} T$.
Therefore,$f_{\max} \propto T$.
If the temperature $T$ is doubled $(T' = 2T)$,the new frequency of maximum intensity becomes $f'_m = 2f_m$.
According to the Stefan-Boltzmann law,the total energy emitted per unit time is $Q = e \sigma A T^4$.
Thus,$Q \propto T^4$.
If the temperature is doubled,the new energy $Q'$ is $Q' \propto (2T)^4 = 16T^4 = 16Q$.
Therefore,the frequency of maximum intensity doubles,and the total emitted energy increases to $16$ times.

(N/A) According to Kirchhoff's law of radiation,good absorbers of radiation are also good emitters. Since a body with high reflectivity reflects most of the incident radiation,it is a poor absorber. Therefore,it is also a poor emitter.
$(b)$ Brass is a good conductor of heat,while wood is a poor conductor. When you touch a brass tumbler,heat flows rapidly from your hand to the metal,causing a sensation of cold. With a wooden tray,heat transfer is very slow,so your hand loses less heat,making it feel warmer.
$(c)$ An optical pyrometer measures temperature based on the intensity of radiation. $A$ red-hot iron piece in the open has an emissivity less than $1$. Since it is not a perfect black body,it emits less radiation than a black body at the same temperature,leading to a lower temperature reading. In a furnace,the environment acts like a cavity (black body),so the iron piece effectively behaves as a black body,yielding an accurate reading.
$(d)$ The atmosphere acts as a blanket for the Earth,trapping infrared radiation emitted by the surface through the greenhouse effect. Without the atmosphere,this heat would escape into space,making the Earth's surface temperature drop drastically.
$(e)$ Steam at $100 \, ^{\circ}C$ contains significantly more energy than water at $100 \, ^{\circ}C$ due to the latent heat of vaporization $(540 \, cal/g)$. As steam condenses in the radiators,it releases this large amount of latent heat,making it more efficient for heating.

(N/A) Thermal energy can be transferred through three distinct mechanisms:
$1$. Thermal Conduction: Heat transfer in solids where particles vibrate about their mean positions without net displacement. It does not require gravity and does not involve the formation of currents.
$2$. Thermal Convection: Heat transfer in fluids (liquids and gases) where particles physically move from one region to another,carrying thermal energy. This process requires gravity and leads to the formation of convection currents.
$3$. Thermal Radiation: Heat transfer in the form of electromagnetic waves (specifically infrared radiation). It does not require any material medium for propagation and can occur in a vacuum.

(D) Heat transfer occurs whenever there is a temperature difference between two objects or regions. The three primary modes of heat transfer are:
$1$. Conduction: The transfer of heat through direct contact between particles of a solid or stationary fluid.
$2$. Convection: The transfer of heat by the actual movement of the heated fluid (liquid or gas) particles.
$3$. Radiation: The transfer of heat through electromagnetic waves,which does not require a material medium.
Since all three processes are driven by temperature differences,the correct answer is $D$.

$(a)$ False. Conduction does not stop; rather, the temperature gradient becomes constant, and the rate of heat transfer remains steady.
$(b)$ True. According to Kirchhoff's law of radiation, for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.
$(c)$ False. $A$ perfect black body is defined by its ability to absorb all incident radiation, regardless of its appearance or color in the visible spectrum.
$(d)$ False. Heat capacity depends on the state of the matter (e.g., phase) and the conditions under which heat is added (e.g., constant pressure vs. constant volume).

(N/A) Given $1 \text{ cal} = 4.184 \text{ J}$ and $1 \text{ cm} = 10^{-2} \text{ m}$.
$0.49 \frac{\text{cal}}{\text{cm} \cdot \text{K} \cdot \text{s}} = 0.49 \times \frac{4.184 \text{ J}}{10^{-2} \text{ m} \cdot \text{K} \cdot \text{s}} = 0.49 \times 418.4 \approx 205 \frac{\text{J}}{\text{m} \cdot \text{K} \cdot \text{s}}$.
Note: If using $1 \text{ cal} = 4.2 \text{ J}$,then $0.49 \times 420 = 205.8 \approx 206$.
$(b)$ If the rate of absorption is greater than the rate of emission,the substance gains net energy,so its temperature will increase.
$(c)$ According to Newton's Law of Cooling,the rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings,provided the difference is small.

(A) $(a) - (iii, iv)$ and $(b) - (ii, iv)$.
$(a)$ $A$ cup of hot tea loses heat to the surroundings primarily through conduction (through the cup material to the air) and radiation (from the surface of the tea and cup).
$(b)$ $A$ substance kept near a fire receives heat primarily through natural convection (as the air near the fire heats up and rises) and radiation (direct thermal radiation from the fire).

(B) The entropy change for a system or body is given by $\Delta S = \frac{Q}{T}$.
For the system to be in a steady state,the heat lost by the reservoir at temperature $T_{1}$ must equal the heat gained by the reservoir at temperature $T_{2}$ (let this be $Q$).
The total rate of change of entropy for the conduction process is the sum of the entropy changes of the two reservoirs:
$\frac{dS}{dt} = \frac{d}{dt} \left( \frac{-Q}{T_{1}} + \frac{Q}{T_{2}} \right) = \left( \frac{dQ}{dt} \right) \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)$
From Fourier's law of heat conduction,the rate of heat flow is $\frac{dQ}{dt} = kA \frac{(T_{1} - T_{2})}{L}$,where $k$ is thermal conductivity,$A$ is the cross-sectional area,and $L$ is the length.
Substituting this into the entropy rate equation:
$\frac{dS}{dt} = \left( kA \frac{(T_{1} - T_{2})}{L} \right) \left( \frac{T_{1} - T_{2}}{T_{1} T_{2}} \right) = \frac{kA}{L} \frac{(T_{1} - T_{2})^2}{T_{1} T_{2}}$
We can rewrite this in terms of the ratio $x = T_{1} / T_{2}$:
$\frac{dS}{dt} = \frac{kA}{L} \frac{T_{2}^2 (x - 1)^2}{T_{2}^2 x} = \frac{kA}{L} \frac{(x - 1)^2}{x} = \frac{kA}{L} \left( x - 2 + \frac{1}{x} \right)$
When $x = 1$ $(T_{1} = T_{2})$,$\frac{dS}{dt} = 0$. For $x > 1$ or $x < 1$,$\frac{dS}{dt} > 0$. The function $f(x) = \frac{(x-1)^2}{x}$ is non-negative and symmetric in its behavior relative to the minimum at $x=1$. Plot $(b)$ correctly represents this relationship.

(C) Let $Q$ be the total incident thermal energy,$Q_a$ be the absorbed energy,$Q_t$ be the transmitted energy,and $Q_r$ be the reflected energy.
Given: $Q = 120 \ J$,$Q_t = 12 \ J$,and coefficient of absorption $a = 0.6$.
We know that the coefficient of absorption $a = \frac{Q_a}{Q}$.
Therefore,$Q_a = a \times Q = 0.6 \times 120 \ J = 72 \ J$.
According to the law of conservation of energy,the total incident energy is the sum of absorbed,transmitted,and reflected energy:
$Q = Q_a + Q_t + Q_r$
$120 \ J = 72 \ J + 12 \ J + Q_r$
$120 \ J = 84 \ J + Q_r$
$Q_r = 120 \ J - 84 \ J = 36 \ J$.
Thus,the amount of heat reflected is $36 \ J$.

(C) According to Wien's displacement law,$\lambda_m T = \text{constant}$.
Given $\lambda_{m1} = \lambda_0$ and $\lambda_{m2} = \frac{\lambda_0}{2}$.
Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2 \implies \lambda_0 T_1 = \frac{\lambda_0}{2} T_2 \implies T_2 = 2T_1$.
According to the Stefan-Boltzmann law,the power radiated by a black body is $P = \sigma A T^4$,which implies $P \propto T^4$.
Thus,the ratio of the new power $P_2$ to the initial power $P_1$ is $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$.
Substituting $T_2 = 2T_1$,we get $\frac{P_2}{P_1} = (2)^4 = 16$.
Hence,the power radiated increases by a factor of $16$.

(B) According to Wien's displacement law,the temperature $T$ is inversely proportional to the wavelength of maximum emission: $T \propto \frac{1}{\lambda_{\max }}$.
From Stefan-Boltzmann law,the power radiated $P$ by a blackbody is given by $P = \sigma A T^4$,where $A$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.
Since $A = \pi r^2$ for a disc (assuming radiation from both sides,$A = 2\pi r^2$),we have $P \propto r^2 T^4$.
Substituting $T \propto \frac{1}{\lambda_{\max }}$,we get $P \propto \frac{r^2}{\lambda_{\max }^4}$.
Given radii $r_x = 2 \ m, r_y = 2 \ m, r_z = 6 \ m$ and wavelengths $\lambda_x = 3 \ \mu m, \lambda_y = 4 \ \mu m, \lambda_z = 5 \ \mu m$:
$P_x \propto \frac{2^2}{3^4} = \frac{4}{81} \approx 0.049$
$P_y \propto \frac{2^2}{4^4} = \frac{4}{256} = 0.0156$
$P_z \propto \frac{6^2}{5^4} = \frac{36}{625} = 0.0576$
Comparing the values,$P_z > P_x > P_y$. Since the question asks for the maximum power,$P_z$ is the maximum.

(B) According to Wien's displacement law,$\lambda_m T = b$ (constant),so $T \propto 1/\lambda_m$.
Given the ratio of wavelengths $\lambda_P : \lambda_Q = 3:4$,the ratio of temperatures is $T_P : T_Q = 4:3$.
According to Stefan-Boltzmann law,the power radiated by a black body is $P = \sigma A T^4 = \sigma (4\pi r^2) T^4$.
Thus,the ratio of power radiated is $\frac{P_P}{P_Q} = \left( \frac{r_P}{r_Q} \right)^2 \left( \frac{T_P}{T_Q} \right)^4$.
Substituting the given values: $\frac{P_P}{P_Q} = \left( \frac{3}{2} \right)^2 \times \left( \frac{4}{3} \right)^4$.
$\frac{P_P}{P_Q} = \frac{9}{4} \times \frac{256}{81} = \frac{1}{1} \times \frac{64}{9} = \frac{64}{9}$.

(B) According to Wien's displacement law, $\lambda_{\max} T = \text{constant}$, so $\lambda_{1} T_{1} = \lambda_{2} T_{2}$.
Given $\lambda_{1} = \lambda$ and $\lambda_{2} = \frac{2 \lambda}{3}$.
Thus, $T_{2} = \frac{\lambda_{1} T_{1}}{\lambda_{2}} = \frac{\lambda T_{1}}{2 \lambda / 3} = \frac{3}{2} T_{1}$.
According to Stefan-Boltzmann law, the emissive power $E$ is proportional to $T^{4}$, i.e., $E = \sigma A T^{4}$.
Therefore, $\frac{E_{2}}{E_{1}} = \left( \frac{T_{2}}{T_{1}} \right)^{4}$.
Substituting the value of $T_{2}$, we get $\frac{E_{2}}{E} = \left( \frac{3/2 T_{1}}{T_{1}} \right)^{4} = \left( \frac{3}{2} \right)^{4} = \frac{81}{16}$.
Hence, $E_{2} = \frac{81 E}{16}$.

(D) The dimension of Stefan's constant $\sigma$ is given by the formula $P = \sigma A T^4$,where $P$ is power,$A$ is area,and $T$ is temperature.
$[\sigma] = \frac{[P]}{[A][T]^4} = \frac{[ML^2 T^{-3}]}{[L^2][K^4]} = [MT^{-3} K^{-4}]$.
The dimension of Wien's constant $b$ is given by $\lambda_{max} T = b$,where $\lambda$ is wavelength and $T$ is temperature.
$[b] = [L][K]$.
Therefore,the dimensions of $\sigma b$ are:
$[\sigma b] = [MT^{-3} K^{-4}] \times [LK] = [L^1 M^1 T^{-3} K^{-3}]$.

(A) Let the thermal conductivity of the rod be $K$ and the cross-sectional area be $A$. The rate of heat flow is given by $H = \frac{KA \Delta T}{L}$.
For the section in ice (length $x_1 = 60 \, cm$): The temperature difference is $\Delta T_1 = 325^{\circ} C - 0^{\circ} C = 325^{\circ} C$. The rate of heat flow is $H_1 = \frac{KA(325)}{60}$.
This heat melts the ice: $H_1 = m_{ice} L_f$, where $L_f$ is the latent heat of fusion.
For the section in boiling water (length $x_2 = 100 - 60 = 40 \, cm$): The temperature difference is $\Delta T_2 = 325^{\circ} C - 100^{\circ} C = 225^{\circ} C$. However, heat flows from $325^{\circ} C$ to $100^{\circ} C$, so $H_2 = \frac{KA(325 - 100)}{40} = \frac{KA(225)}{40}$.
This heat converts water to steam: $H_2 = m_{steam} L_v$, where $L_v = 6.75 L_f$.
Given $m_{steam} = 2 \, g/s$, then $H_2 = 2 \times 6.75 L_f = 13.5 L_f$.
Equating $H_2$: $\frac{KA(225)}{40} = 13.5 L_f$ implies $KA = \frac{13.5 \times 40}{225} L_f = 2.4 L_f$.
Now substitute $KA$ into $H_1$: $H_1 = \frac{2.4 L_f \times 325}{60} = 0.04 \times 325 L_f = 13 L_f$.
Since $H_1 = m_{ice} L_f$, we get $m_{ice} = 13 \, g/s$.

(A) Statement $A$ is true: Convection is the process of heat transfer by the actual movement of fluid particles due to density differences caused by unequal temperatures.
Statement $B$ is true: When a hot bar is placed under running tap water,the water in contact with the bar heats up,becomes less dense,and rises,while cooler water replaces it. This circulation is convection.
Statement $C$ is true: Heat is defined as energy in transit due to a temperature difference. Without a temperature gradient,there is no net heat transfer.
Therefore,all three statements are correct.

(D) Given that the rate of heat loss by convection is equal to the rate of heat loss by radiation.
Let $h$ be the convection coefficient,$A$ be the surface area,$e$ be the emissivity,$\sigma$ be the Stefan-Boltzmann constant,$T$ be the body temperature,and $T_0$ be the surrounding temperature.
$T = 400 + 273 = 673 \ K$
$T_0 = 30 + 273 = 303 \ K$
The rate of heat loss by convection is $P_{conv} = hA(T - T_0)$.
The rate of heat loss by radiation is $P_{rad} = eA\sigma(T^4 - T_0^4)$.
Equating the two: $hA(T - T_0) = eA\sigma(T^4 - T_0^4)$.
$20(673 - 303) = e(5.67 \times 10^{-8})(673^4 - 303^4)$.
$20(370) = e(5.67 \times 10^{-8})(2.049 \times 10^{11} - 0.0084 \times 10^{11})$.
$7400 = e(5.67 \times 10^{-8})(2.0406 \times 10^{11})$.
$7400 = e(11570.2)$.
$e = 7400 / 11570.2 \approx 0.6397$.
Rounding to the nearest option,$e \approx 0.66$.

(D) According to Stefan-Boltzmann law,the power radiated by a black body is $P \propto T^4$.
According to Wien's displacement law,the wavelength corresponding to maximum energy is $\lambda \propto \frac{1}{T}$,which implies $T \propto \frac{1}{\lambda}$.
Substituting this into the power relation,we get $P \propto \left(\frac{1}{\lambda}\right)^4 = \frac{1}{\lambda^4}$.
Therefore,the ratio of power radiated is $\frac{P_1}{P_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^4$.
Given $\lambda_2 = \frac{\lambda_1}{2}$,we have $\frac{\lambda_2}{\lambda_1} = \frac{1}{2}$.
Substituting this value,$\frac{P_1}{P_2} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
Thus,$P_2 = 16 P_1 = 16 P$.

(B) According to Stefan's law,the energy $(E)$ emitted per unit area per unit time is given by $E = \sigma T^4$,where $\sigma$ is Stefan's constant and $T$ is temperature.
Dimensional analysis for $\sigma$:
$\frac{[M L^2 T^{-2}]}{[L^2 T]} = [\sigma] [K^4]$
$[M T^{-3}] = [\sigma] [K^4]$
$[\sigma] = [M T^{-3} K^{-4}]$
Wien's displacement law is given by $b = \lambda T$,where $\lambda$ is wavelength and $T$ is temperature.
Dimension of $b = [L K]$.
Now,calculating the dimensions of $\sigma b^4$:
$[\sigma b^4] = [M T^{-3} K^{-4}] \times [L K]^4$
$[\sigma b^4] = [M T^{-3} K^{-4}] \times [L^4 K^4]$
$[\sigma b^4] = [M L^4 T^{-3}]$.