Heat Conduction and Thermal Conductivity Questions in English

(D) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series must be the same.
The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{L}$, where $K$ is thermal conductivity, $A$ is area, $L$ is thickness, and $(T_1 - T_2)$ is the temperature difference.
Let the temperature of the junction $EF$ be $T$.
For the first slab (thickness $d$, conductivity $2k$): $H_1 = \frac{(2k)A(100 - T)}{d}$
For the second slab (thickness $2d$, conductivity $k$): $H_2 = \frac{kA(T - 0)}{2d}$
Since $H_1 = H_2$:
$\frac{2kA(100 - T)}{d} = \frac{kA(T - 0)}{2d}$
$2(100 - T) = \frac{T}{2}$
$4(100 - T) = T$
$400 - 4T = T$
$5T = 400$
$T = 80^oC$.

(A) In a steady state,the rate of heat flow $(H)$ through both sheets must be the same.
$H = \frac{K_1 A (T_A - T_B)}{d} = \frac{K_2 A (T_B - T_C)}{2d}$
Given that $T_A, T_B, T_C$ are in geometric progression with common ratio $r = 2$,we have:
$T_B = 2 T_A$ and $T_C = 4 T_A$.
Substituting these into the heat flow equation:
$\frac{K_1 (T_A - 2 T_A)}{d} = \frac{K_2 (2 T_A - 4 T_A)}{2d}$
$\frac{K_1 (-T_A)}{d} = \frac{K_2 (-2 T_A)}{2d}$
$- \frac{K_1 T_A}{d} = - \frac{K_2 T_A}{d}$
$K_1 = K_2$
Therefore,the ratio $\frac{K_1}{K_2} = 1$.

(B) In steady state,the rate of heat flow $\dot{Q}$ through both layers $A$ and $B$ must be the same.
Given: Thickness $d_A = d_B = d$,Thermal conductivity $k_A = 2k$ and $k_B = k$.
The rate of heat flow is given by $\dot{Q} = \frac{kA \Delta T}{d}$.
Since $\dot{Q}$ is constant and $A$ and $d$ are the same for both layers,we have $k_A \Delta T_A = k_B \Delta T_B$.
Substituting the values: $(2k) \Delta T_A = (k) \Delta T_B$.
Given $\Delta T_B = 36^{\circ}C$,we get $2 \Delta T_A = 36^{\circ}C$.
Therefore,$\Delta T_A = \frac{36}{2} = 18^{\circ}C$.

(B) Since there is no heat flow in rod $AB$,the temperature at junction $B$ must be equal to the temperature at $A$.
Therefore,$\theta_B = \theta_A = 20^{\circ} C$.
Since the heat flowing from $D$ to $B$ must be equal to the heat flowing from $B$ to $C$ (as no heat flows into $AB$),we have:
$\frac{KA(90^{\circ} - 20^{\circ})}{L_{BD}} = \frac{KA(20^{\circ} - 0^{\circ})}{L_{BC}}$
$\frac{70}{L_{BD}} = \frac{20}{L_{BC}}$
$\frac{L_{BD}}{L_{BC}} = \frac{70}{20} = \frac{7}{2}$
Thus,the ratio of the lengths of $BD$ to $BC$ is $7/2$.

(C) Let $\theta$ be the temperature of the junction and $H_{1}, H_{2}$,and $H_{3}$ be the heat currents flowing through the rods.
According to the principle of conservation of energy at the junction,the sum of heat currents entering the junction must equal the sum of heat currents leaving it.
Assuming heat flows from the $30^{\circ}C$ end towards the junction,and from the junction towards the $20^{\circ}C$ and $10^{\circ}C$ ends:
$H_{1} = H_{2} + H_{3}$
Using the formula for heat current $H = \frac{KA(\Delta T)}{L}$,where $K$ is thermal conductivity,$A$ is cross-sectional area,and $L$ is length:
$\frac{30 - \theta}{\left(\frac{30}{KA}\right)} = \frac{\theta - 20}{\left(\frac{20}{KA}\right)} + \frac{\theta - 10}{\left(\frac{10}{KA}\right)}$
Canceling $KA$ from both sides:
$\frac{30 - \theta}{30} = \frac{\theta - 20}{20} + \frac{\theta - 10}{10}$
Multiply by $60$ to clear the denominators:
$2(30 - \theta) = 3(\theta - 20) + 6(\theta - 10)$
$60 - 2\theta = 3\theta - 60 + 6\theta - 60$
$60 - 2\theta = 9\theta - 120$
$11\theta = 180$
$\theta = \frac{180}{11} \approx 16.36^{\circ}C \approx 16.4^{\circ}C$.

(B) The rate of heat transfer $H$ is given by $H = \frac{K A \Delta T}{l}$.
Given: $A = 9.0 \, cm^2 = 9.0 \times 10^{-4} \, m^2$,$l = 0.54 \, m$,$\Delta T = 100 \, ^\circ C - 0 \, ^\circ C = 100 \, K$.
The rate of heat flow is also equal to the rate of melting of ice: $H = \frac{m L_f}{t}$,where $L_f = 334 \, J/g$ (latent heat of fusion of ice).
$H = \frac{1 \, g \times 334 \, J/g}{33 \, s} \approx 10.12 \, W$.
Equating the two expressions: $\frac{K \times 9.0 \times 10^{-4} \times 100}{0.54} = \frac{334}{33}$.
$K \times \frac{0.09}{0.54} = 10.12$.
$K \times \frac{1}{6} = 10.12$.
$K = 60.72 \, W m^{-1} K^{-1}$.
Rounding to the nearest integer,$K = 60 \, W m^{-1} K^{-1}$.

(A) For a steady-state heat flow through a series combination of layers,the rate of heat flow $\frac{dQ}{dt}$ is constant through both layers.
The rate of heat flow is given by $\frac{dQ}{dt} = k A \frac{\Delta T}{\Delta x}$,where $k$ is the thermal conductivity,$A$ is the cross-sectional area,and $\frac{\Delta T}{\Delta x}$ is the temperature gradient (slope of the $T-x$ graph).
Since $\frac{dQ}{dt}$ and $A$ are constant,we have $\frac{\Delta T}{\Delta x} \propto \frac{1}{k}$.
From the graph,the slope of the temperature profile in layer $(1)$ is less than the slope in layer $(2)$,i.e.,$|\frac{\Delta T}{\Delta x}|_1 < |\frac{\Delta T}{\Delta x}|_2$.
Since the slope is inversely proportional to $k$,a smaller slope implies a larger thermal conductivity.
Therefore,$k_1 > k_2$.

(B) The rate of heat flow through a rod is given by: $\dot{Q} = \frac{KA \Delta T}{l}$.
Since the rods are connected in series,the rate of heat flow $\dot{Q}$ through both rods must be the same.
For rod $A$ (thermal conductivity $3K$): $\dot{Q}_A = \frac{(3K)A T_A}{l}$.
For rod $B$ (thermal conductivity $K$): $\dot{Q}_B = \frac{KA T_B}{l}$.
Since $\dot{Q}_A = \dot{Q}_B$,we have:
$\frac{3KA T_A}{l} = \frac{KA T_B}{l}$
$3 T_A = T_B$
Therefore,$\frac{T_A}{T_B} = \frac{1}{3}$.

(B) In a series connection,the rate of heat flow $(H)$ through both rods is the same.
The rate of heat flow is given by $H = KA \frac{\Delta T}{l}$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,and $\Delta T/l$ is the temperature gradient $(G)$.
Since $H$ is constant for both rods,we have $H = K_A A G_A = K_B A G_B$.
Given $K_A = 3K$ and $K_B = K$,we get $(3K) A G_A = (K) A G_B$.
Simplifying this,we get $3 G_A = G_B$.
Therefore,the ratio of the temperature gradients is $\frac{G_A}{G_B} = \frac{1}{3}$.

(A) Let $K_1$ and $K_2$ be the thermal conductivities of the thinner sheet (thickness $d$) and the thicker sheet (thickness $3d$) respectively.
In steady state,the rate of heat flow through both sheets is the same:
$\frac{dQ}{dt} = \frac{K_1 A (A - B)}{d} = \frac{K_2 A (B - C)}{3d}$
Since $A, B, C$ are in arithmetic progression,we have $B - A = C - B$,which implies $A - B = B - C$.
Substituting this into the heat flow equation:
$\frac{K_1}{d} = \frac{K_2}{3d}$
$\frac{K_1}{K_2} = \frac{1}{3}$
Thus,the ratio of the thermal conductivity of the thinner sheet to the thicker sheet is $1 : 3$.

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.
Since the rod is insulated,$H$ is constant along the length.
Substituting $K = \frac{\alpha}{T}$,we get $H = -\frac{\alpha A}{T} \frac{dT}{dx}$.
Rearranging the terms: $\frac{H}{\alpha A} dx = -\frac{dT}{T}$.
Integrating from $x=0$ to $x$ and $T=T_1$ to $T$:
$\int_0^x \frac{H}{\alpha A} dx = -\int_{T_1}^T \frac{dT}{T}$.
$\frac{H}{\alpha A} x = -(\ln T - \ln T_1) = \ln \frac{T_1}{T}$.
At $x=L$,$T=T_2$,so $\frac{H}{\alpha A} L = \ln \frac{T_1}{T_2}$.
Thus,$\frac{H}{\alpha A} = \frac{1}{L} \ln \frac{T_1}{T_2}$.
Substituting this back: $\frac{x}{L} \ln \frac{T_1}{T_2} = \ln \frac{T_1}{T}$.
$\ln \left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \ln \frac{T_1}{T}$.
Taking the exponential of both sides: $\left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \frac{T_1}{T}$.
Therefore,$T = T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$.

(A) In steady state,the rate of heat flow through both sections is the same.
Let $T$ be the temperature at the interface.
The rate of heat flow $H$ is given by $H = \frac{KA(T_{high} - T_{low})}{l}$.
For the first section: $H = \frac{K_1 A(T_1 - T)}{l_1}$.
For the second section: $H = \frac{K_2 A(T - T_2)}{l_2}$.
Equating the two rates: $\frac{K_1 A(T_1 - T)}{l_1} = \frac{K_2 A(T - T_2)}{l_2}$.
$\frac{K_1(T_1 - T)}{l_1} = \frac{K_2(T - T_2)}{l_2}$.
$K_1 l_2(T_1 - T) = K_2 l_1(T - T_2)$.
$K_1 l_2 T_1 - K_1 l_2 T = K_2 l_1 T - K_2 l_1 T_2$.
$K_1 l_2 T_1 + K_2 l_1 T_2 = T(K_1 l_2 + K_2 l_1)$.
$T = \frac{K_1 l_2 T_1 + K_2 l_1 T_2}{K_1 l_2 + K_2 l_1}$.

(C) In a steady state,the rate of heat flow through both sections of the rod must be equal.
Let $T$ be the temperature at the interface.
The rate of heat flow $H$ is given by $H = \frac{KA(T_{high} - T_{low})}{l}$.
Since the cross-sectional area $A$ is the same for both sections,we have:
$\frac{K_1 A (T_1 - T)}{l_1} = \frac{K_2 A (T - T_2)}{l_2}$
$\frac{K_1 (T_1 - T)}{l_1} = \frac{K_2 (T - T_2)}{l_2}$
$K_1 l_2 (T_1 - T) = K_2 l_1 (T - T_2)$
$K_1 l_2 T_1 - K_1 l_2 T = K_2 l_1 T - K_2 l_1 T_2$
$K_1 l_2 T_1 + K_2 l_1 T_2 = T (K_1 l_2 + K_2 l_1)$
$T = \frac{K_1 l_2 T_1 + K_2 l_1 T_2}{K_1 l_2 + K_2 l_1}$
Thus,the correct option is $C$.

(D) In the steady-state,the rate of heat flow $H$ through any cross-section of a uniform metallic bar is constant.
According to Fourier's law of heat conduction,$H = -kA \frac{d\theta}{dx}$,where $k$ is the thermal conductivity,$A$ is the cross-sectional area,and $\frac{d\theta}{dx}$ is the temperature gradient.
Since $H$,$k$,and $A$ are constants for a uniform bar,the temperature gradient $\frac{d\theta}{dx} = -\frac{H}{kA}$ must also be a constant.
Integrating this expression with respect to $x$,we get $\theta(x) = -(\frac{H}{kA})x + C$,where $C$ is a constant of integration representing the temperature at the hot end $(x=0)$.
This equation is of the form $\theta = -mx + C$,which represents a straight line with a negative slope.
Therefore,the temperature decreases linearly along the length of the bar.

(C) Let the temperature of junction $D$ be $T_D$. In steady state, the net heat current at junction $D$ is zero: $i_{AD} + i_{CD} = 0$.
$\frac{KA}{L}(100 - T_D) + \frac{KA}{L}(0 - T_D) = 0 \implies 100 - T_D - T_D = 0 \implies 2T_D = 100 \implies T_D = 50^{\circ}C$. (Statement $B$ is correct).
Now, calculate heat currents:
$i_{AB} = \frac{KA}{L}(100 - 40) = 1 \times 60 = 60 \text{ J/s}$.
$i_{BC} = \frac{KA}{L}(40 - 0) = 1 \times 40 = 40 \text{ J/s}$.
$i_{AD} = \frac{KA}{L}(100 - 50) = 50 \text{ J/s}$.
$i_{DC} = \frac{KA}{L}(50 - 0) = 50 \text{ J/s}$.
Comparing $i_{AB}$ and $i_{BC}$: $i_{AB} = 60 \text{ J/s}$ and $i_{BC} = 40 \text{ J/s}$. Thus, $i_{AB} = 1.5 \times i_{BC}$. (Statement $A$ is correct, Statement $C$ is incorrect).
Heat current withdrawn at $B$: The junction $B$ receives heat from $A$ $(i_{AB} = 60 \text{ J/s})$ and loses heat to $C$ $(i_{BC} = 40 \text{ J/s})$. The remaining heat must be withdrawn from $B$: $i_{withdrawn} = i_{AB} - i_{BC} = 60 - 40 = 20 \text{ J/s}$. (Statement $D$ is correct).

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

(B) Let the temperature of the junction be $T$.
Since the rods are connected at a junction,the net heat flow into the junction must be zero (steady state condition).
The rate of heat flow through a rod is given by $H = \frac{KA(T_1 - T_2)}{L}$.
Let $A$ be the cross-sectional area and $L$ be the length of each rod. The thermal resistance of each rod is $R = \frac{L}{KA}$.
For the three rods,the heat currents are:
$H_1 = \frac{K A (110 - T)}{L}$
$H_2 = \frac{2K A (20 - T)}{L}$
$H_3 = \frac{3K A (T - 0)}{L}$
According to the principle of conservation of energy at the junction,$H_1 + H_2 + H_3 = 0$.
$\frac{KA}{L} (110 - T) + \frac{2KA}{L} (20 - T) + \frac{3KA}{L} (T - 0) = 0$
Dividing by $\frac{KA}{L}$,we get:
$(110 - T) + 2(20 - T) + 3T = 0$
$110 - T + 40 - 2T + 3T = 0$
$150 = 0$
Wait,re-evaluating the heat flow direction: The heat flows from higher temperature to lower temperature. Let $T$ be the junction temperature.
$H_1 = \frac{KA(110 - T)}{L}$
$H_2 = \frac{2KA(20 - T)}{L}$
$H_3 = \frac{3KA(T - 0)}{L}$
Sum of heat currents leaving the junction is zero: $H_1 + H_2 + H_3 = 0$ is incorrect if we assume all are leaving. Let's assume all heat flows towards the junction or away.
Actually,the sum of heat currents flowing into the junction is zero: $\frac{KA(110-T)}{L} + \frac{2KA(20-T)}{L} = \frac{3KA(T-0)}{L}$.
$110 - T + 2(20 - T) = 3T$
$110 - T + 40 - 2T = 3T$
$150 - 3T = 3T$
$6T = 150$
$T = 25\ ^oC$.

(D) The rate of heat flow $(H)$ through a rod is given by the formula $H = \frac{KA \Delta T}{L}$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $L$ is the length of the rod.
Since the material is the same,$K$ is constant. Given that $\Delta T$ is also the same,we have $H \propto \frac{A}{L}$.
The area $A = \pi r^2 = \pi (d/2)^2$,so $A \propto d^2$.
Thus,the ratio of heat flow is $\frac{H_1}{H_2} = \left( \frac{d_1}{d_2} \right)^2 \times \left( \frac{L_2}{L_1} \right)$.
Given $\frac{d_1}{d_2} = \frac{1}{2}$ and $\frac{L_1}{L_2} = \frac{2}{1}$,we substitute these values:
$\frac{H_1}{H_2} = \left( \frac{1}{2} \right)^2 \times \left( \frac{1}{2} \right) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Therefore,the ratio is $1:8$.

(B) In steady state,the rate of heat flow $(H)$ through both layers $A$ and $B$ is the same.
$H = \frac{K_A A \Delta T_A}{L} = \frac{K_B A \Delta T_B}{L}$
Since the area $A$ and thickness $L$ are the same for both layers,we have:
$K_A \Delta T_A = K_B \Delta T_B$
Given that the thermal conductivity of $A$ is twice that of $B$,i.e.,$K_A = 2K_B$:
$2K_B \Delta T_A = K_B \Delta T_B \implies \Delta T_B = 2 \Delta T_A$
The total temperature difference across the wall is given as $36\,^{\circ}C$:
$\Delta T_A + \Delta T_B = 36$
Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:
$\Delta T_A + 2 \Delta T_A = 36$
$3 \Delta T_A = 36$
$\Delta T_A = 12\,^{\circ}C$

(C) In a steady state,the rate of heat flow through both bars must be equal.
Let the interface temperature be $\varphi$.
The rate of heat flow $H$ is given by $H = \frac{KA(\Delta T)}{L}$.
For the first bar: $H_1 = \frac{K \cdot A \cdot (\varphi - 0)}{1}$.
For the second bar: $H_2 = \frac{3K \cdot A \cdot (100 - \varphi)}{2}$.
Since $H_1 = H_2$,we have:
$K \cdot A \cdot \varphi = \frac{3K \cdot A \cdot (100 - \varphi)}{2}$.
Canceling $K$ and $A$ from both sides:
$\varphi = \frac{3(100 - \varphi)}{2}$.
$2\varphi = 300 - 3\varphi$.
$5\varphi = 300$.
$\varphi = 60\,^{\circ}C$.

(B) In a steady state,the rate of heat flow through rods in series is the same,i.e.,$\left(\frac{dQ}{dt}\right)_{A} = \left(\frac{dQ}{dt}\right)_{B}$.
Using the formula $\frac{dQ}{dt} = \frac{KA \Delta T}{L}$,where $K$ is thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $L$ is the length.
For rod $A$: $L_A = 30 \text{ cm}$,$\Delta T_A = 100 - 70 = 30^{\circ}\text{C}$.
For rod $B$: $L_B = 100 - 30 = 70 \text{ cm}$,$\Delta T_B = 70 - 35 = 35^{\circ}\text{C}$.
Assuming the cross-sectional area $A$ is the same for both rods:
$\frac{K_A A (30)}{30} = \frac{K_B A (35)}{70}$
$K_A = K_B \times \frac{35}{70}$
$K_A = K_B \times 0.5$
Therefore,$\frac{K_A}{K_B} = 0.5$.

(B) In a steady state,the rate of heat flow $(H)$ through both rods connected in series is the same.
$H_A = H_B$
$\frac{K_A A \Delta T_A}{L_A} = \frac{K_B A \Delta T_B}{L_B}$
Given: $L_A = 30 \text{ cm}$,$L_B = 70 \text{ cm}$,$\Delta T_A = 100 - 70 = 30 \text{ } ^\circ\text{C}$,$\Delta T_B = 70 - 35 = 35 \text{ } ^\circ\text{C}$.
Assuming the cross-sectional area $A$ is the same for both rods:
$\frac{K_A \cdot 30}{30} = \frac{K_B \cdot 35}{70}$
$K_A = K_B \cdot 0.5$
$\frac{K_A}{K_B} = 0.5$

(C) Let $\theta$ be the temperature of the junction,and $H_1, H_2, H_3$ be the heat currents flowing through the rods.
According to the principle of conservation of energy at the junction,the sum of heat currents entering the junction must equal the sum of heat currents leaving the junction.
From the diagram,$H_1 = H_2 + H_3$.
The heat current $H$ is given by $H = \frac{KA(\Delta T)}{L}$,where $K$ is thermal conductivity,$A$ is cross-sectional area,and $L$ is length.
Given that $K$ and $A$ are the same for all rods,we have:
$\frac{30 - \theta}{(30/KA)} = \frac{\theta - 20}{(20/KA)} + \frac{\theta - 10}{(10/KA)}$
Multiplying by $KA$,we get:
$\frac{30 - \theta}{30} = \frac{\theta - 20}{20} + \frac{\theta - 10}{10}$
Multiplying the entire equation by $60$:
$2(30 - \theta) = 3(\theta - 20) + 6(\theta - 10)$
$60 - 2\theta = 3\theta - 60 + 6\theta - 60$
$60 - 2\theta = 9\theta - 120$
$11\theta = 180$
$\theta = \frac{180}{11} \approx 16.36^{\circ}C \approx 16.4^{\circ}C$.

(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.
The rate of heat flow through a rod is given by $H = \frac{KA(T_1 - T_2)}{L}$.
For the rod with conductivity $3k$ connected to $100\,^{\circ}C$,the heat flow towards the junction is $H_1 = \frac{3k A(100 - \theta)}{L}$.
For the other two rods,the heat flows away from the junction towards $50\,^{\circ}C$ and $0\,^{\circ}C$ respectively:
$H_2 = \frac{2k A(\theta - 50)}{L}$
$H_3 = \frac{k A(\theta - 0)}{L}$
By the junction rule,$H_1 = H_2 + H_3$:
$\frac{3k A(100 - \theta)}{L} = \frac{2k A(\theta - 50)}{L} + \frac{k A(\theta - 0)}{L}$
Canceling $\frac{kA}{L}$ from both sides:
$3(100 - \theta) = 2(\theta - 50) + \theta$
$300 - 3\theta = 2\theta - 100 + \theta$
$300 - 3\theta = 3\theta - 100$
$400 = 6\theta$
$\theta = \frac{400}{6} = \frac{200}{3}\,^{\circ}C$

(B) The heat current $H$ through a rod is given by $H = -KA \frac{dT}{dx}$,where $A$ is the cross-sectional area and $K$ is the thermal conductivity.
Given $K = C/T$,we have $H = -\frac{C}{T} A \frac{dT}{dx}$.
Rearranging the terms,we get $H \frac{dx}{A} = -C \frac{dT}{T}$.
Integrating both sides from $x=0$ to $x=l$ and $T=T_1$ to $T=T_2$:
$\int_0^l \frac{H}{A} dx = -\int_{T_1}^{T_2} C \frac{dT}{T}$.
Since the rod is in steady state,the heat current density $J_H = H/A$ is constant.
$J_H \int_0^l dx = -C [\ln T]_{T_1}^{T_2}$.
$J_H \cdot l = -C (\ln T_2 - \ln T_1) = C \ln \left( \frac{T_1}{T_2} \right)$.
Assuming the magnitude of the heat current density (taking $T_1 > T_2$ for flow),we get $J_H = \frac{C}{l} \ln \left( \frac{T_1}{T_2} \right)$.
Note: The options provided in the prompt contain a sign discrepancy; based on standard convention,the magnitude is $\frac{C}{l} \ln \left( \frac{T_1}{T_2} \right)$.

(B) In a steady state,the rate of heat flow through any cross-section of the rod is constant.
Let $\theta$ be the temperature at point $C$,which is at a distance of $x = 0.6\,m$ from end $A$.
The rate of heat flow is given by $\frac{dQ}{dt} = \frac{KA(\theta_1 - \theta_2)}{L}$.
For the entire rod $AB$ (length $L = 1\,m$): $\frac{dQ}{dt} = \frac{KA(80 - 0)}{1} = 80KA$.
For the segment $AC$ (length $x = 0.6\,m$): $\frac{dQ}{dt} = \frac{KA(80 - \theta)}{0.6}$.
Equating the two rates: $80KA = \frac{KA(80 - \theta)}{0.6}$.
$80 = \frac{80 - \theta}{0.6}$.
$80 \times 0.6 = 80 - \theta$.
$48 = 80 - \theta$.
$\theta = 80 - 48 = 32^{\circ}C$.

(A) Let the thermal conductivity of steel be $K_{steel} = k$.
Then,the thermal conductivity of copper is $K_{copper} = 9k$.
Let the temperature of the junction be $\theta$.
Since the rods are in series and insulated from the surroundings,the rate of heat flow $(H)$ through both rods must be equal in steady state.
$H = \frac{K_{copper} A (100 - \theta)}{L/2} = \frac{K_{steel} A (\theta - 0)}{L/2}$
Since the cross-sectional area $A$ and length $L/2$ are the same for both:
$K_{copper} (100 - \theta) = K_{steel} (\theta - 0)$
$9k (100 - \theta) = k \theta$
$900 - 9\theta = \theta$
$10\theta = 900$
$\theta = 90\,^oC$.

(A) The energy flux (heat current density) $J$ is given by the formula: $J = \frac{1}{A} \frac{dQ}{dt} = k \frac{\Delta T}{\ell}$.
Given values are:
Thermal conductivity $k = 0.1\, W\, K^{-1}\, m^{-1}$.
Temperature difference $\Delta T = T_1 - T_2 = 10^3 - 10^2 = 1000 - 100 = 900\, K$.
Thickness $\ell = 1\, m$.
Substituting these values into the formula:
$J = 0.1 \times \frac{900}{1} = 90\, W\, m^{-2}$.
Therefore,the energy flux is $90\, W\, m^{-2}$.

(B) At steady state,the rate of heat flow through both materials is the same.
Let $\theta$ be the temperature at the interface.
The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.
For the first material: $H_1 = \frac{(3K)A(\theta_2 - \theta)}{d}$.
For the second material: $H_2 = \frac{KA(\theta - \theta_1)}{3d}$.
Equating $H_1$ and $H_2$:
$\frac{3KA(\theta_2 - \theta)}{d} = \frac{KA(\theta - \theta_1)}{3d}$
$9(\theta_2 - \theta) = \theta - \theta_1$
$9\theta_2 - 9\theta = \theta - \theta_1$
$10\theta = 9\theta_2 + \theta_1$
$\theta = \frac{9\theta_2 + \theta_1}{10} = \frac{\theta_1}{10} + \frac{9\theta_2}{10}$.

(B) The rate of heat flow $(Q)$ through a rod is given by the formula: $Q = \frac{KA \Delta T}{L}$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $L$ is the length.
Since $K$ and $\Delta T$ are constant for all rods,$Q \propto \frac{A}{L}$.
Given $A = \pi r^2$,we have $Q \propto \frac{r^2}{L}$.
We calculate the ratio $\frac{r^2}{L}$ for each option:
$A: \frac{1^2}{50} = \frac{1}{50} = 0.02$
$B: \frac{2^2}{100} = \frac{4}{100} = 0.04$
$C: \frac{0.5^2}{25} = \frac{0.25}{25} = 0.01$
$D: \frac{1.5^2}{75} = \frac{2.25}{75} = 0.03$
Comparing these values,the ratio is maximum for option $B$ $(0.04)$. Therefore,the rod in option $B$ conducts the most heat.

(B) In a steady state,the rate of heat flow $H$ is constant throughout the rod.
$H = \frac{T_A - T_B}{R_{AB}} = \frac{T_A - T_C}{R_{AC}}$
Since the thermal resistance $R$ is proportional to the length $L$ of the rod $(R = \frac{L}{kA})$,we have $\frac{R_{AC}}{R_{AB}} = \frac{L_{AC}}{L_{AB}}$.
Therefore,$\frac{T_A - T_C}{T_A - T_B} = \frac{L_{AC}}{L_{AB}}$.
Substituting the given values: $T_A = 100\,^{\circ}C$,$T_B = 0\,^{\circ}C$,$L_{AC} = 9\,cm$,and $L_{AB} = 20\,cm$.
$\frac{100 - T_C}{100 - 0} = \frac{9}{20}$
$100 - T_C = \frac{9}{20} \times 100$
$100 - T_C = 45$
$T_C = 100 - 45 = 55\,^{\circ}C$.

(C) In steady state,the heat current $H$ flowing through both rods is the same.
The heat current is given by $H = \frac{KA \Delta T}{L}$,which can be written as $\Delta T = H \cdot \frac{L}{KA}$.
For a small segment of length $dx$,the temperature drop is $dT = H \cdot \frac{dx}{KA}$.
Thus,the temperature gradient is $\frac{dT}{dx} = -\frac{H}{KA}$.
Since $H$ and $A$ are constant,the magnitude of the temperature gradient is inversely proportional to the thermal conductivity $K$ (i.e.,$|\frac{dT}{dx}| \propto \frac{1}{K}$).
Given $K_{Cu} > K_{Steel}$,the magnitude of the temperature gradient in the copper rod is smaller than that in the steel rod.
This means the slope of the $T-x$ graph is less steep for the copper rod $(0 < x < L)$ and steeper for the steel rod $(L < x < 2L)$.
Graph $C$ correctly represents this behavior.

(A) In steady-state heat conduction through rods in series,the rate of heat flow $\frac{dH}{dt}$ is the same for both rods.
The formula for heat flow is $\frac{dH}{dt} = K A \left| \frac{dT}{dx} \right|$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,and $\left| \frac{dT}{dx} \right|$ is the temperature gradient.
Since $\frac{dH}{dt}$ and $A$ are constant for both rods,we have $K_1 \left| \frac{dT}{dx} \right|_1 = K_2 \left| \frac{dT}{dx} \right|_2$.
From the given graph,the slope of the temperature-distance line represents the magnitude of the temperature gradient $\left| \frac{dT}{dx} \right|$.
It is clear from the graph that the slope for the second rod (from $L$ to $2L$) is greater than the slope for the first rod (from $0$ to $L$).
Therefore,$\left| \frac{dT}{dx} \right|_2 > \left| \frac{dT}{dx} \right|_1$.
Since $K_1 \left| \frac{dT}{dx} \right|_1 = K_2 \left| \frac{dT}{dx} \right|_2$,and $\left| \frac{dT}{dx} \right|_2 > \left| \frac{dT}{dx} \right|_1$,it must be that $K_1 > K_2$.

(B) Let the temperature of the junction be $T_0$.
According to the principle of steady-state heat flow,the sum of heat currents flowing into the junction must be zero:
$\frac{T_0 - 100}{R_1} + \frac{T_0 - 50}{R_2} + \frac{T_0 - 0}{R_3} = 0$
Since the rods have the same dimensions (length $L$ and area $A$),the thermal resistance $R = \frac{L}{kA}$.
Thus,$R_1 = \frac{L}{3KA}$,$R_2 = \frac{L}{2KA}$,and $R_3 = \frac{L}{KA}$.
Substituting these values into the equation:
$\frac{T_0 - 100}{L/(3KA)} + \frac{T_0 - 50}{L/(2KA)} + \frac{T_0 - 0}{L/(KA)} = 0$
$3K' (T_0 - 100) + 2K' (T_0 - 50) + K' (T_0) = 0$ (where $K' = \frac{KA}{L}$)
$3T_0 - 300 + 2T_0 - 100 + T_0 = 0$
$6T_0 = 400$
$T_0 = \frac{400}{6} = \frac{200}{3}^{\circ}C$

(C) In a steady state,the rate of heat flow through both parts of the rod must be equal.
Let $K$ be the thermal conductivity of the metal,$l_1 = l_2 = 1\, m$ be the lengths of the two parts,and $T$ be the temperature at the junction $C$.
The rate of heat flow is given by $H = \frac{KA \Delta T}{l}$.
For the first part: $H_1 = \frac{K(2A)(100 - T)}{1}$.
For the second part: $H_2 = \frac{K(A)(T - 70)}{1}$.
Since $H_1 = H_2$,we have:
$2KA(100 - T) = KA(T - 70)$
$2(100 - T) = T - 70$
$200 - 2T = T - 70$
$3T = 270$
$T = 90\,^{\circ}C$.

(B) In the steady state,the rate of heat flow through both rods must be equal.
Let $T$ be the temperature of the interface.
The rate of heat flow is given by $\frac{dQ}{dt} = \frac{kA(T_H - T_L)}{\ell}$.
For the rod with thermal conductivity $2k$ (connected to $100^oC$): $\frac{dQ_1}{dt} = \frac{2kA(100 - T)}{\ell}$.
For the rod with thermal conductivity $k$ (connected to $20^oC$): $\frac{dQ_2}{dt} = \frac{kA(T - 20)}{\ell}$.
Equating the two rates: $\frac{2kA(100 - T)}{\ell} = \frac{kA(T - 20)}{\ell}$.
$2(100 - T) = T - 20$.
$200 - 2T = T - 20$.
$3T = 220$.
$T = \frac{220}{3} \approx 73.3^oC$.

(B) cooking utensil should have:
$1$. High thermal conductivity so that it can conduct heat efficiently from the source to the food inside.
$2$. Low specific heat capacity so that the utensil reaches the required temperature quickly with minimal heat absorption by the material itself.

(C) The rate of heat conduction through a rod is given by $\frac{Q}{t} = \frac{kA(T_1 - T_2)}{l}$,where $A = \pi r^2$ is the cross-sectional area and $l$ is the length.
When the rod is melted and reshaped,the volume $V = A \cdot l = \pi r^2 l$ remains constant.
Let the new radius be $r' = \frac{r}{2}$.
Since $V = \pi (r')^2 l' = \pi (\frac{r}{2})^2 l' = \pi \frac{r^2}{4} l'$,we have $\pi r^2 l = \pi \frac{r^2}{4} l'$,which gives the new length $l' = 4l$.
The new area is $A' = \pi (r')^2 = \pi (\frac{r}{2})^2 = \frac{A}{4}$.
The heat conducted by the new rod in time $t$ is $Q' = \frac{kA' (T_1 - T_2) t}{l'}$.
Substituting $A' = \frac{A}{4}$ and $l' = 4l$:
$Q' = \frac{k (A/4) (T_1 - T_2) t}{4l} = \frac{1}{16} \frac{kA(T_1 - T_2) t}{l}$.
Since $Q = \frac{kA(T_1 - T_2) t}{l}$,we get $Q' = \frac{Q}{16}$.

(A) For a conical rod of length $L$ with radii $r_1$ and $r_2$ at the ends,the radius at a distance $x$ from the end with radius $r_1$ is given by $r(x) = r_1 + \frac{r_2 - r_1}{L}x$.
The thermal resistance $dR$ of a small element of thickness $dx$ is $dR = \frac{dx}{K A(x)} = \frac{dx}{K \pi (r(x))^2}$.
The total thermal resistance $R$ is the integral of $dR$ from $0$ to $L$:
$R = \int_{0}^{L} \frac{dx}{K \pi (r_1 + \frac{r_2 - r_1}{L}x)^2}$.
Solving this integral gives $R = \frac{L}{K \pi r_1 r_2}$.
The rate of heat flow is $\frac{dQ}{dt} = \frac{T_2 - T_1}{R} = \frac{K \pi r_1 r_2 (T_2 - T_1)}{L}$.

(B) The rate of heat flow is given by the formula $\frac{dQ}{dt} = \frac{KA\Delta T}{L}$.
For the first rod (connected to $100\,^{\circ}C$): $\left( \frac{dQ}{dt} \right)_1 = \frac{3KA}{L}(100 - \theta)$.
For the second rod (connected to $50\,^{\circ}C$): $\left( \frac{dQ}{dt} \right)_2 = \frac{2KA}{L}(\theta - 50)$.
For the third rod (connected to $20\,^{\circ}C$): $\left( \frac{dQ}{dt} \right)_3 = \frac{KA}{L}(\theta - 20)$.
At the junction,by the principle of conservation of energy (steady state),the heat flowing in must equal the heat flowing out:
$\left( \frac{dQ}{dt} \right)_1 = \left( \frac{dQ}{dt} \right)_2 + \left( \frac{dQ}{dt} \right)_3$.
Substituting the expressions:
$\frac{3KA}{L}(100 - \theta) = \frac{2KA}{L}(\theta - 50) + \frac{KA}{L}(\theta - 20)$.
Canceling $\frac{KA}{L}$ from both sides:
$3(100 - \theta) = 2(\theta - 50) + (\theta - 20)$.
$300 - 3\theta = 2\theta - 100 + \theta - 20$.
$300 - 3\theta = 3\theta - 120$.
$6\theta = 420$.
$\theta = 70\,^{\circ}C$.

(A) Brass is a metal and a good conductor of heat. On a cold day,when a brass tumbler is touched,heat transfers rapidly from our body to the brass due to its high thermal conductivity. Since our body loses heat quickly,the tumbler feels cold.
On the other hand,wood is a poor conductor of heat (insulator). The transfer of heat from our body to the wood is very slow and minimal,hence the wooden tray does not feel as cold as the brass tumbler.

(D) The rate of heat flow is given by the formula: $\frac{dQ}{dt} = -K A \frac{dT}{dx}$.
Here,$\frac{dQ}{dt}$ is the heat flow rate in Watts ($W$ or $J/s$),$A$ is the area in $m^2$,and $\frac{dT}{dx}$ is the temperature gradient in $K/m$.
Rearranging for thermal conductivity $(K)$:
$K = \frac{(dQ/dt)}{A \cdot (dT/dx)}$
Substituting the units:
$K = \frac{W}{m^2 \cdot (K/m)} = \frac{W}{m \cdot K} = W m^{-1} K^{-1}$.
Therefore,the unit of thermal conductivity is $W m^{-1} K^{-1}$.

(A) The heat conducted through the ice layer of thickness $x$ in time $dt$ is given by $dQ = \frac{KA(T_2 - T_1)}{x} dt$.
Here,$T_2 = 0^{\circ}C$ (temperature at the water-ice interface) and $T_1 = -26^{\circ}C$ (outside air temperature).
So,$dQ = \frac{KA(0 - (-26))}{x} dt = \frac{26KA}{x} dt$.
This heat causes an additional thickness $dx$ of water to freeze. The mass of this new ice layer is $dm = A \cdot dx \cdot \rho$.
The heat released during this phase change is $dQ = dm \cdot L = A \cdot dx \cdot \rho \cdot L$.
Equating the two expressions for $dQ$:
$\frac{26KA}{x} dt = A \cdot dx \cdot \rho \cdot L$.
Rearranging to find the rate of increase of thickness $\frac{dx}{dt}$:
$\frac{dx}{dt} = \frac{26K}{\rho x L}$.

(B) In the steady state,the rate of heat flow through the steel rod must equal the rate of heat flow through the copper rod.
Let $T$ be the temperature of the junction.
The rate of heat flow is given by $H = \frac{KA(T_H - T_L)}{L}$.
For the steel rod $(1)$: $K_1 = 50.2 \; J s^{-1} m^{-1} K^{-1}$,$A_1 = 2A_2$,$L_1 = 0.15 \; m$,$T_H = 300^{\circ} C$,$T_L = T$.
For the copper rod $(2)$: $K_2 = 385 \; J s^{-1} m^{-1} K^{-1}$,$A_2$,$L_2 = 0.10 \; m$,$T_H = T$,$T_L = 0^{\circ} C$.
Equating the heat flow rates:
$\frac{K_1 A_1 (300 - T)}{L_1} = \frac{K_2 A_2 (T - 0)}{L_2}$
$\frac{50.2 \times (2 A_2) \times (300 - T)}{0.15} = \frac{385 \times A_2 \times T}{0.10}$
$\frac{100.4 \times (300 - T)}{0.15} = \frac{385 \times T}{0.10}$
$669.33 \times (300 - T) = 3850 \times T$
$200799 - 669.33 T = 3850 T$
$4519.33 T = 200799$
$T \approx 44.4^{\circ} C$.

(A) Side of the cubical ice box,$s = 30 \,cm = 0.3 \,m$.
Thickness of the ice box,$l = 5.0 \,cm = 0.05 \,m$.
Mass of ice initially,$m_{initial} = 4 \,kg$.
Time duration,$t = 6 \,h = 6 \times 3600 \,s = 21600 \,s$.
Temperature difference,$\Delta T = 45 \,^{\circ}C - 0 \,^{\circ}C = 45 \,K$.
Coefficient of thermal conductivity,$K = 0.01 \,J \,s^{-1} \,m^{-1} \,K^{-1}$.
Latent heat of fusion of water,$L = 335 \times 10^{3} \,J \,kg^{-1}$.
Surface area of the cubical box,$A = 6s^{2} = 6 \times (0.3)^{2} = 0.54 \,m^{2}$.
The heat conducted into the box is given by $Q = \frac{K A \Delta T t}{l}$.
$Q = \frac{0.01 \times 0.54 \times 45 \times 21600}{0.05} = 104976 \,J$.
The mass of ice that melts,$m_{melt} = \frac{Q}{L} = \frac{104976}{335 \times 10^{3}} \approx 0.313 \,kg$.
The remaining mass of ice is $m_{remaining} = m_{initial} - m_{melt} = 4 - 0.313 = 3.687 \,kg \approx 3.69 \,kg$.

(B) Thickness of the boiler,$l = 1.0\; cm = 0.01\; m$.
Boiling rate of water,$R = 6.0\; kg/min$.
Mass,$m = 6.0\; kg$.
Time,$t = 1\; min = 60\; s$.
Thermal conductivity of brass,$K = 109\; J s^{-1} m^{-1} K^{-1}$.
Heat of vaporisation,$L = 2256 \times 10^{3}\; J kg^{-1}$.
The amount of heat flowing into water through the brass base of the boiler is given by:
$Q = \frac{K A (T_{1} - T_{2}) t}{l} \dots (i)$
Where $T_{1}$ is the temperature of the flame and $T_{2} = 100^{\circ}C$ is the boiling point of water.
Heat required for boiling the water is $Q = m L \dots (ii)$.
Equating $(i)$ and $(ii)$:
$m L = \frac{K A (T_{1} - T_{2}) t}{l}$
$T_{1} - T_{2} = \frac{m L l}{K A t} = \frac{6.0 \times 2256 \times 10^{3} \times 0.01}{109 \times 0.15 \times 60} \approx 137.98^{\circ}C$.
Thus,$T_{1} = 137.98 + 100 = 237.98^{\circ}C \approx 238^{\circ}C$.

(N/A) Consider a metallic bar of length $L$ and uniform cross-sectional area $A$ with its two ends maintained at different temperatures. This can be done by placing the ends in thermal contact with large reservoirs at temperatures $T_{C}$ and $T_{D}$ respectively,where $T_{C} > T_{D}$.
Assuming the ideal condition that the sides of the bar are fully insulated,no heat is exchanged with the surroundings. Initially,the temperatures of different parts of the bar increase with time. After some time,a 'steady state' is reached where the temperature of the bar decreases uniformly with distance from $T_{C}$ to $T_{D}$.
In this steady state,the reservoir at $C$ supplies heat at a constant rate,which is transferred through the bar and given out at the same rate to the reservoir at $D$. Both the rate of heat flow $\frac{dQ}{dt}$ and the temperature gradient $\frac{dT}{dx}$ remain constant with time.
Experimentally,it is found that the rate of heat flow (or heat current) $H$ is given by:
$H = \frac{dQ}{dt} = KA \frac{T_{C} - T_{D}}{L}$
where $K$ is the thermal conductivity of the material.

(N/A) Copper is an excellent conductor of heat. By coating the bottom of a cooking vessel with copper,the heat from the stove is distributed more rapidly and uniformly across the base of the pot. This prevents the formation of hot spots and ensures that food is cooked evenly. Therefore,copper coating is used to improve the thermal efficiency of cooking vessels.

(N/A) Thermal conductivity $(k)$ is defined as the rate of heat transfer through a unit thickness of a material per unit area per unit temperature difference.
It is given by the formula: $Q = \frac{kA(T_1 - T_2)t}{d}$,where $Q$ is the heat transferred,$A$ is the area,$d$ is the thickness,and $(T_1 - T_2)$ is the temperature difference.
Rearranging for $k$: $k = \frac{Qd}{A(T_1 - T_2)t}$.
$SI$ Unit: The unit of heat is Joule $(J)$,area is $m^2$,thickness is $m$,temperature is Kelvin $(K)$,and time is second $(s)$. Thus,the unit is $\frac{J \cdot m}{m^2 \cdot K \cdot s} = W \cdot m^{-1} \cdot K^{-1}$.
Dimensional Formula: Since $Q$ is energy $([ML^2T^{-2}])$,$d$ is length $([L])$,$A$ is area $([L^2])$,$\Delta T$ is temperature $([K])$,and $t$ is time $([T])$:
$k = \frac{[ML^2T^{-2}][L]}{[L^2][K][T]} = [MLT^{-3}K^{-1}]$.

(A) Thermal conductivity $(k)$ is a material property that depends on the nature of the substance. For a material to have a constant thermal conductivity,it must be homogeneous (uniform composition throughout) and isotropic (having the same physical properties in all directions). If these conditions are met,the thermal conductivity remains independent of the temperature gradient or the geometry of the object.