Thermal Resistance and it's Combination Questions in English

(C) Let the temperature of the junction be $\theta$. Since the rods have the same material,cross-section,and length,they have the same thermal resistance $R$.
The rate of heat flow through a rod is given by $H = \frac{\Delta T}{R}$.
At the junction,the sum of the heat flow rates must be zero (steady state):
$\frac{90 - \theta}{R} + \frac{60 - \theta}{R} + \frac{30 - \theta}{R} + \frac{0 - \theta}{R} = 0$
Multiplying by $R$:
$(90 - \theta) + (60 - \theta) + (30 - \theta) + (0 - \theta) = 0$
$180 - 4\theta = 0$
$4\theta = 180$
$\theta = 45^{\circ}C$.

(C) For arrangement $(a)$,the rods are in series. Let the length of each rod be $L$ and the cross-sectional area be $A$. The thermal resistance of each rod is $R = \frac{L}{KA}$.
The effective resistance in series is $R_{\text{eff}, a} = R_{Al} + R_{Cu} + R_{Al} = \frac{L}{A} \left( \frac{1}{K_{Al}} + \frac{1}{K_{Cu}} + \frac{1}{K_{Al}} \right) = \frac{L}{A} \left( \frac{2}{200} + \frac{1}{400} \right) = \frac{L}{A} \left( \frac{4+1}{400} \right) = \frac{5L}{400A} = \frac{L}{80A}$.
The rate of heat flow is $H_1 = \frac{\Delta T}{R_{\text{eff}, a}} = \frac{100 \Delta T}{L/A} = 40 \, W$.
For arrangement $(b)$,the rods are in parallel. The effective resistance is given by $\frac{1}{R_{\text{eff}, b}} = \frac{1}{R_{Al}} + \frac{1}{R_{Cu}} + \frac{1}{R_{Al}} = \frac{A}{L} (K_{Al} + K_{Cu} + K_{Al}) = \frac{A}{L} (200 + 400 + 200) = \frac{800A}{L}$.
Thus,$R_{\text{eff}, b} = \frac{L}{800A}$.
The rate of heat flow is $H_2 = \frac{\Delta T}{R_{\text{eff}, b}} = \frac{\Delta T}{L/(800A)} = \frac{800A \Delta T}{L}$.
From the first case,$\frac{A \Delta T}{L} = \frac{40}{100/80} = \frac{40 \times 80}{100} = 32$.
Substituting this into $H_2$: $H_2 = 800 \times 32 = 25600$ (Wait,re-evaluating: $H_1 = \frac{\Delta T}{L/(80A)} = 80 \frac{A \Delta T}{L} = 40 \implies \frac{A \Delta T}{L} = 0.5$).
Then $H_2 = 800 \times 0.5 = 400 \, W$.

(B) Let the temperature of the junction $O$ be $\theta$.
According to the principle of steady-state heat flow,the sum of heat currents flowing into the junction must equal the sum of heat currents flowing out of the junction.
Heat current $H = \frac{KA(\Delta T)}{L}$.
Assuming heat flows from the $30^{\circ} C$ end towards the junction and then out to the $20^{\circ} C$ and $10^{\circ} C$ ends:
$\frac{KA}{30}(30 - \theta) = \frac{KA}{20}(\theta - 20) + \frac{KA}{10}(\theta - 10)$
Dividing both sides by $KA$:
$\frac{30 - \theta}{30} = \frac{\theta - 20}{20} + \frac{\theta - 10}{10}$
Multiply by $60$ to simplify:
$2(30 - \theta) = 3(\theta - 20) + 6(\theta - 10)$
$60 - 2\theta = 3\theta - 60 + 6\theta - 60$
$60 - 2\theta = 9\theta - 120$
$180 = 11\theta$
$\theta = \frac{180}{11} \approx 16.36^{\circ} C$.
Rounding to one decimal place,we get $16.4^{\circ} C$.

(A) Let the four corners of the square be $A, B, C,$ and $D$ in cyclic order. Let the temperatures at $A$ and $C$ be $T_A = 40^{\circ} C$ and $T_C = 10^{\circ} C$.
Since the four rods are identical,each rod acts as a thermal resistance $R = \frac{L}{kA}$.
The arrangement of the four rods forms a Wheatstone bridge circuit where the heat flows from $A$ to $C$ through two parallel paths: $A \rightarrow B \rightarrow C$ and $A \rightarrow D \rightarrow C$.
Due to the symmetry of the circuit and the identical nature of the rods,the potential (temperature) at $B$ and $D$ must be equal.
Since $T_B = T_D$,the temperature difference between the other two corners $B$ and $D$ is $T_B - T_D = 0^{\circ} C$.

(C) The thermal resistance of a slab is given by $R = \frac{L}{kA}$,where $L$ is the length,$k$ is thermal conductivity,and $A$ is the cross-sectional area. Let $b$ be the width of the slabs.
$R_A = \frac{L}{2K(4Lb)} = \frac{1}{8} \frac{L}{K b} = \frac{R_0}{8}$
$R_B = \frac{4L}{3K(Lb)} = \frac{4}{3} \frac{L}{K b} = \frac{4R_0}{3}$
$R_C = \frac{4L}{4K(2Lb)} = \frac{1}{2} \frac{L}{K b} = \frac{R_0}{2}$
$R_D = \frac{4L}{5K(Lb)} = \frac{4}{5} \frac{L}{K b} = \frac{4R_0}{5}$
$R_E = \frac{L}{6K(4Lb)} = \frac{1}{24} \frac{L}{K b} = \frac{R_0}{24}$
$(i)$ Since slabs $A$ and $E$ are in series with the parallel combination of $B, C, D$,the total heat current $Q$ passes through both $A$ and $E$. Thus,heat flow through $A$ and $E$ is the same. Statement $(A)$ is correct.
$(ii)$ The temperature difference across a slab is $\Delta T = Q \cdot R$. Since $R_E$ is the smallest resistance,the temperature difference across $E$ is the smallest. Statement $(C)$ is correct.
$(iii)$ For the parallel section,the heat current $Q$ splits into $i_B, i_C, i_D$. Since they are in parallel,the potential difference $\Delta T_{BCD}$ is the same for all. Thus $i = \frac{\Delta T}{R}$.
$i_C = \frac{\Delta T}{R_C} = \frac{2\Delta T}{R_0}$,$i_B = \frac{\Delta T}{R_B} = \frac{3\Delta T}{4R_0}$,$i_D = \frac{\Delta T}{R_D} = \frac{5\Delta T}{4R_0}$.
Summing $i_B + i_D = \frac{3\Delta T}{4R_0} + \frac{5\Delta T}{4R_0} = \frac{8\Delta T}{4R_0} = \frac{2\Delta T}{R_0} = i_C$. Statement $(D)$ is correct.
Therefore,$(A, C, D)$ are correct.

(A) Let the length of each block be $L$,cross-sectional area be $A$,and thermal resistance of the block with conductivity $k$ be $R = \frac{L}{kA}$. Then the resistance of the block with conductivity $2k$ is $R' = \frac{L}{2kA} = \frac{R}{2}$.
In configuration $I$ (series): The blocks are in series. The equivalent thermal resistance is $R_{eq,I} = R + \frac{R}{2} = \frac{3R}{2}$.
The heat current $H_I = \frac{\Delta T}{R_{eq,I}} = \frac{2\Delta T}{3R}$.
In configuration $II$ (parallel): The blocks are in parallel. The equivalent thermal resistance is given by $\frac{1}{R_{eq,II}} = \frac{1}{R} + \frac{1}{R/2} = \frac{1}{R} + \frac{2}{R} = \frac{3}{R}$. So,$R_{eq,II} = \frac{R}{3}$.
The heat current $H_{II} = \frac{\Delta T}{R_{eq,II}} = \frac{3\Delta T}{R}$.
Since the amount of heat $Q$ transported is the same,$Q = H_I t_I = H_{II} t_II$.
$t_{II} = t_I \times \frac{H_I}{H_{II}} = 9 \times \frac{2\Delta T / 3R}{3\Delta T / R} = 9 \times \frac{2}{9} = 2.0 \ s$.

(C) Let the length of each conductor be $L$ and the area of cross-section of the $1^{\text{st}}$ and $2^{\text{nd}}$ conductors be $A$. Then the area of the $3^{\text{rd}}$ conductor is $2A$.
Thermal resistance is given by $R = \frac{L}{kA}$.
$R_1 = \frac{L}{k_1 A} = \frac{L}{60A}$,$R_2 = \frac{L}{k_2 A} = \frac{L}{120A}$,$R_3 = \frac{L}{k_3 (2A)} = \frac{L}{135 \times 2A} = \frac{L}{270A}$.
Conductors $1$ and $2$ are in parallel,so their equivalent resistance $R_{12}$ is:
$\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{60A}{L} + \frac{120A}{L} = \frac{180A}{L} \implies R_{12} = \frac{L}{180A}$.
In steady state,the heat current $H$ flowing through the combination is constant:
$H = \frac{100 - \theta}{R_{12}} = \frac{\theta - 0}{R_3}$
$\frac{100 - \theta}{L / 180A} = \frac{\theta}{L / 270A}$
$180(100 - \theta) = 270\theta$
$2(100 - \theta) = 3\theta$
$200 - 2\theta = 3\theta$
$5\theta = 200 \implies \theta = 40^{\circ}C$.

(C) The thermal resistance $R$ of a rod is given by $R = \frac{L}{KA}$,where $L$ is the length,$K$ is the thermal conductivity,and $A = \pi r^2$ is the cross-sectional area.
For the two rods in series,the rate of heat flow $\frac{dQ}{dt}$ is the same in steady state:
$\frac{dQ}{dt} = \frac{\Delta T}{R} = \frac{400 - T}{R_A} = \frac{T - 200}{R_B}$
$\frac{400 - T}{T - 200} = \frac{R_A}{R_B} = \left( \frac{L_A}{L_B} \right) \left( \frac{K_B}{K_A} \right) \left( \frac{r_B}{r_A} \right)^2$
Given $\frac{L_A}{L_B} = \frac{1}{2}$,$\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$,and $\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$.
Substituting these values:
$\frac{R_A}{R_B} = \left( \frac{1}{2} \right) \times (2) \times \left( \frac{1}{2} \right)^2 = 1 \times \frac{1}{4} = \frac{1}{4}$
Now,$\frac{400 - T}{T - 200} = \frac{1}{4}$
$4(400 - T) = T - 200$
$1600 - 4T = T - 200$
$5T = 1800$
$T = 360 \ K$

(B) The total heat current $I = 150 \ J/s$ enters the junction. The path splits into two parallel branches: the straight path and the bent path.
Let the length of the straight path be $L_2 = 60 \ cm$ and the length of the bent path be $L_1 = 30 + 60 + 30 = 120 \ cm$.
Since thermal resistance $R = \frac{L}{kA}$,and assuming the cross-sectional area $A$ and thermal conductivity $k$ are uniform,$R \propto L$.
Thus,$R_2 \propto 60$ and $R_1 \propto 120$.
The heat current $I_1$ through the bent path is given by the current divider rule for thermal circuits:
$I_1 = I \times \frac{R_2}{R_1 + R_2} = 150 \times \frac{60}{120 + 60} = 150 \times \frac{60}{180} = 150 \times \frac{1}{3} = 50 \ J/s$.

(D) The system can be modeled as an equivalent thermal circuit. Each rod has a thermal resistance $R$.
Between points $B$ and $C$,there are two parallel paths,each consisting of two rods in series. The resistance of each path is $R + R = 2R$.
The equivalent resistance $R_{BC}$ of these two parallel paths is given by $\frac{1}{R_{BC}} = \frac{1}{2R} + \frac{1}{2R} = \frac{2}{2R} = \frac{1}{R}$,so $R_{BC} = R$.
The total thermal resistance of the circuit from $A$ to $D$ is $R_{AD} = R_{AB} + R_{BC} + R_{CD} = R + R + R = 3R$.
The steady-state heat current $H$ flowing through the circuit is $H = \frac{\Delta T}{R_{total}} = \frac{100 - 10}{3R} = \frac{90}{3R} = \frac{30}{R}$.
The temperature at junction $C$ can be found using the heat current through the rod $CD$:
$H = \frac{T_C - 10}{R}$
$\frac{30}{R} = \frac{T_C - 10}{R}$
$30 = T_C - 10$
$T_C = 40^{\circ} C$.

(B) Let the length of each rod be $\ell$,area of cross-section be $A$,and thermal conductivity be $K$. The heat transferred is given by $Q = \frac{KA \Delta T}{L_{eq}} \times t$.
Case $1$: Series combination (end to end).
The equivalent length is $L_{eq} = \ell + \ell = 2\ell$. The equivalent area is $A$.
$Q = \frac{KA \Delta T}{2\ell} \times 12 \quad \dots (1)$
Case $2$: Parallel combination (along their lengths).
The equivalent length is $L_{eq} = \ell$. The equivalent area is $A_{eq} = A + A = 2A$.
$Q = \frac{K(2A) \Delta T}{\ell} \times t \quad \dots (2)$
Since the heat $Q$ and temperature difference $\Delta T$ are the same in both cases,we equate $(1)$ and $(2)$:
$\frac{KA \Delta T}{2\ell} \times 12 = \frac{2KA \Delta T}{\ell} \times t$
$\frac{12}{2} = 2t$
$6 = 2t$
$t = 3 \ s$.

(B) Let the heat transferred be $Q$.
When rods are joined end to end,the equivalent thermal resistance is $R_{eq} = R + R = 2R$,where $R = \frac{l}{KA}$.
The rate of heat transfer is $\frac{Q}{t_1} = \frac{\Delta \theta}{R_{eq}} = \frac{\Delta \theta}{2 \frac{l}{KA}} = \frac{KA \Delta \theta}{2l}$.
Given $t_1 = 12 \ s$,so $Q = \frac{KA \Delta \theta}{2l} \times 12 \dots (i)$.
When rods are joined lengthwise (parallel),the equivalent thermal resistance is $R'_{eq} = \frac{R \times R}{R + R} = \frac{R}{2}$,where $R = \frac{l}{KA}$.
The rate of heat transfer is $\frac{Q}{t_2} = \frac{\Delta \theta}{R'_{eq}} = \frac{\Delta \theta}{R/2} = \frac{2KA \Delta \theta}{l}$.
So,$Q = \frac{2KA \Delta \theta}{l} \times t_2 \dots (ii)$.
Equating $(i)$ and $(ii)$:
$\frac{2KA \Delta \theta}{l} \times t_2 = \frac{KA \Delta \theta}{2l} \times 12$
$2 t_2 = \frac{12}{2}$
$4 t_2 = 12$
$t_2 = 3 \ s$.

(D) In a steady state,the rate of heat flow $(H)$ through materials in series is the same.
For the first material: $H = \frac{KA(T_2 - T)}{x}$,where $T$ is the interface temperature.
For the second material: $H = \frac{(2K)A(T - T_1)}{4x} = \frac{KA(T - T_1)}{2x}$.
Equating the two expressions: $\frac{KA(T_2 - T)}{x} = \frac{KA(T - T_1)}{2x}$.
$2(T_2 - T) = T - T_1 \implies 2T_2 - 2T = T - T_1 \implies 3T = 2T_2 + T_1 \implies T = \frac{2T_2 + T_1}{3}$.
Substituting $T$ back into the first expression: $H = \frac{KA}{x} (T_2 - \frac{2T_2 + T_1}{3}) = \frac{KA}{x} (\frac{3T_2 - 2T_2 - T_1}{3}) = \frac{KA(T_2 - T_1)}{3x}$.
Comparing this with the given expression $\left[\frac{A(T_2 - T_1)K}{x}\right] f$,we find $f = 1/3$.

(A) The thermal resistance $R$ of a rod is given by the formula $R = \frac{l}{KA}$,where $l$ is the length,$K$ is the thermal conductivity,and $A$ is the area of cross-section.
Given that the area of cross-section is the same for both rods,$A_A = A_B = A$.
Given the ratio of thermal conductivities is $\frac{K_A}{K_B} = \frac{3}{2}$.
Since the thermal resistances are equal,$R_A = R_B$.
Substituting the formula,we get $\frac{l_A}{K_A A} = \frac{l_B}{K_B A}$.
Simplifying this,we get $\frac{l_A}{l_B} = \frac{K_A}{K_B}$.
Substituting the given ratio,$\frac{l_A}{l_B} = \frac{3}{2}$.

(A) The heat current through the first slab is given by $\dot{Q}_1 = \frac{K_1 A (T_1 - T)}{d_1}$.
For the second slab,the heat current is $\dot{Q}_2 = \frac{K_2 A (T - T_2)}{d_2}$.
Since the slabs are connected in series,the same heat current flows through both,so $\dot{Q}_1 = \dot{Q}_2$.
Equating the two expressions: $\frac{K_1 A (T_1 - T)}{d_1} = \frac{K_2 A (T - T_2)}{d_2}$.
Canceling $A$ from both sides: $\frac{K_1 (T_1 - T)}{d_1} = \frac{K_2 (T - T_2)}{d_2}$.
Cross-multiplying gives: $K_1 d_2 (T_1 - T) = K_2 d_1 (T - T_2)$.
Expanding: $K_1 d_2 T_1 - K_1 d_2 T = K_2 d_1 T - K_2 d_1 T_2$.
Rearranging to solve for $T$: $K_1 d_2 T_1 + K_2 d_1 T_2 = T (K_1 d_2 + K_2 d_1)$.
Thus,$T = \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1}$.

(B) Let $R$ be the thermal resistance of each rod. When connected in series,the total thermal resistance is $R_{s} = R + R = 2R$.
The heat transferred in series is $Q = \frac{\Delta T}{R_{s}} \times t_{s} = \frac{\Delta T}{2R} \times 8 = \frac{4 \Delta T}{R}$.
When connected in parallel,the equivalent thermal resistance is $R_{p} = \frac{R \times R}{R + R} = \frac{R}{2}$.
The heat transferred in parallel is $Q = \frac{\Delta T}{R_{p}} \times t_{p} = \frac{\Delta T}{R/2} \times t_{p} = \frac{2 \Delta T}{R} \times t_{p}$.
Since the heat transferred $Q$ is the same in both cases,we equate the expressions:
$\frac{4 \Delta T}{R} = \frac{2 \Delta T}{R} \times t_{p}$.
Solving for $t_{p}$,we get $t_{p} = \frac{4}{2} = 2 \ s$.

(B) The thermal resistance $(R_{th})$ is defined as the ratio of the temperature difference $(\Delta T)$ to the rate of heat flow ($H$ or thermal current).
Formula: $R_{th} = \frac{\Delta T}{H}$
Given:
Temperature difference $\Delta T = 28^{\circ} C$
Rate of heat flow $H = 1400 \ cal s^{-1}$
Calculation:
$R_{th} = \frac{28}{1400} \ ^{\circ} C s cal^{-1}$
$R_{th} = 0.02 \ ^{\circ} C s cal^{-1}$
Therefore, the thermal resistance is $0.02 \ ^{\circ} C s cal^{-1}$.

(A) Given:
Rate of flow of heat (conduction rate) $P_{\text{cond}} = 1600 \text{ cal/s}$.
Temperature difference $\Delta T = 40^{\circ} C$.
The formula for thermal resistance $R_T$ is given by the ratio of temperature difference to the rate of heat flow:
$R_T = \frac{\Delta T}{P_{\text{cond}}}$
Substituting the given values:
$R_T = \frac{40}{1600}$
$R_T = \frac{1}{40} = 0.025^{\circ} C s/cal$.
Thus,the thermal resistance is $0.025^{\circ} C s/cal$.

(D) The thermal resistance $R$ of a slab is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area.
For the first material: $R_1 = \frac{x}{KA}$.
For the second material: $R_2 = \frac{4x}{(2K)A} = \frac{2x}{KA}$.
Since the slabs are in series,the equivalent thermal resistance $R_{eq}$ is:
$R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}$.
The rate of heat transfer $\frac{dQ}{dt}$ is given by:
$\frac{dQ}{dt} = \frac{T_2 - T_1}{R_{eq}} = \frac{T_2 - T_1}{\frac{3x}{KA}} = \frac{KA(T_2 - T_1)}{3x}$.
Comparing this with the given expression $\left[\frac{A(T_2 - T_1)K}{x}\right] \cdot f$,we get:
$f = \frac{1}{3}$.

(A) Let the thermal resistance of each rod be $R$ and the temperature of the junction be $T_X$.
According to the principle of conservation of energy,the total heat flowing into the junction must equal the total heat flowing out of the junction.
Here,heat flows from the two rods at $90^{\circ} C$ towards the junction $X$,and then from the junction $X$ towards the rod at $0^{\circ} C$.
Let $H_1$ and $H_2$ be the heat currents from the two rods at $90^{\circ} C$,and $H$ be the heat current flowing towards the $0^{\circ} C$ end.
$H = H_1 + H_2$
Using the formula for heat current $H = \frac{\Delta T}{R}$,we have:
$\frac{T_X - 0}{R} = \frac{90 - T_X}{R} + \frac{90 - T_X}{R}$
Since the rods are identical,the thermal resistance $R$ is the same for all.
$T_X = (90 - T_X) + (90 - T_X)$
$T_X = 180 - 2T_X$
$3T_X = 180$
$T_X = 60^{\circ} C$
Thus,the temperature at the junction $X$ is $60^{\circ} C$.

(B) The thermal resistance of a conductor is given by $R_{th} = \frac{L}{KA}$,where $L$ is length,$K$ is thermal conductivity,and $A$ is the cross-sectional area. Since the conductors are identical,$L$ and $A$ are the same for all.
Let $R_0 = \frac{L}{KA}$. Then the thermal resistances are:
$R_A = \frac{L}{KA} = R_0$
$R_B = \frac{L}{(2K)A} = \frac{R_0}{2}$
$R_C = \frac{L}{(K/2)A} = 2R_0$
In steady state,the heat current $H$ through the series combination is constant:
$H = \frac{\Delta T}{R_{total}} = \frac{100 - 0}{R_A + R_B + R_C} = \frac{100}{R_0 + \frac{R_0}{2} + 2R_0} = \frac{100}{3.5 R_0} = \frac{200}{7 R_0}$
Let $T'$ be the temperature of the junction of $A$ and $B$. The heat current through $A$ is:
$H = \frac{100 - T'}{R_A} = \frac{100 - T'}{R_0}$
Equating the two expressions for $H$:
$\frac{100 - T'}{R_0} = \frac{200}{7 R_0}$
$100 - T' = \frac{200}{7}$
$T' = 100 - 28.57 = 71.43^{\circ} C$
Thus,the temperature is nearly $71^{\circ} C$.

(A) Given that the rods are identical,let the length of each rod be $l$ and the cross-sectional area be $A_{area}$.
Let $K_{A}, K_{B}, K_{C}$ be the thermal conductivities of rods $A, B, C$ respectively.
According to the problem:
$K_{B} = \frac{1}{2} K_{A}$
$K_{B} = 3 K_{C} \implies K_{C} = \frac{K_{B}}{3} = \frac{K_{A}}{6}$
Since the rods are connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual resistances:
$R_{eq} = R_{A} + R_{B} + R_{C}$
$\frac{3l}{K_{eq} A_{area}} = \frac{l}{K_{A} A_{area}} + \frac{l}{K_{B} A_{area}} + \frac{l}{K_{C} A_{area}}$
$\frac{3}{K_{eq}} = \frac{1}{K_{A}} + \frac{1}{K_{A}/2} + \frac{1}{K_{A}/6}$
$\frac{3}{K_{eq}} = \frac{1}{K_{A}} + \frac{2}{K_{A}} + \frac{6}{K_{A}}$
$\frac{3}{K_{eq}} = \frac{9}{K_{A}}$
$K_{eq} = \frac{3 K_{A}}{9} = \frac{1}{3} K_{A}$

(C) In the steady state,the rate of heat flow $(H)$ through both layers connected in series is the same.
Let $l$ be the thickness of each layer and $A$ be the cross-sectional area.
Let $K_A$ and $K_B$ be the thermal conductivities of layers $A$ and $B$ respectively.
Given: $K_A = 2K_B$.
The rate of heat flow is given by $H = \frac{K A \Delta T}{l}$.
Since $H$ is constant for both layers:
$H = \frac{K_A A \Delta T_A}{l} = \frac{K_B A \Delta T_B}{l}$
where $\Delta T_A$ and $\Delta T_B$ are the temperature differences across layers $A$ and $B$ respectively.
Canceling $A$ and $l$ from both sides:
$K_A \Delta T_A = K_B \Delta T_B$
Substituting $K_A = 2K_B$:
$2K_B \Delta T_A = K_B \Delta T_B \Rightarrow \Delta T_B = 2 \Delta T_A$
The total temperature difference across the wall is $\Delta T_A + \Delta T_B = 24^{\circ} C$.
Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:
$\Delta T_A + 2 \Delta T_A = 24^{\circ} C$
$3 \Delta T_A = 24^{\circ} C \Rightarrow \Delta T_A = 8^{\circ} C$.
Therefore,the temperature difference across layer $B$ is:
$\Delta T_B = 2 \times 8^{\circ} C = 16^{\circ} C$.

(D) The system consists of three layers in series: glass, air, and glass. The equivalent thermal resistance $R_{eq}$ is given by $R_{eq} = R_1 + R_2 + R_3$, where $R = \frac{l}{kA}$.
Given: $l_1 = l_3 = 1 \,cm = 0.01 \,m$, $l_2 = 5 \,cm = 0.05 \,m$, $A = 2.6 \,m^2$, $k_{glass} = 0.8 \,Wm^{-1} K^{-1}$, $k_{air} = 0.08 \,Wm^{-1} K^{-1}$.
$R_1 = R_3 = \frac{0.01}{0.8 \times 2.6} = \frac{0.01}{2.08} \approx 0.0048 \,K/W$.
$R_2 = \frac{0.05}{0.08 \times 2.6} = \frac{0.05}{0.208} \approx 0.2404 \,K/W$.
$R_{eq} = 2 \times \left(\frac{0.01}{2.08}\right) + \frac{0.05}{0.208} = \frac{0.02}{2.08} + \frac{0.5}{2.08} = \frac{0.52}{2.08} = 0.25 \,K/W$.
The rate of heat flow $H = \frac{\Delta T}{R_{eq}} = \frac{18 - (-2)}{0.25} = \frac{20}{0.25} = 80 \,W$.

(B) Let the temperature at junction $B$ be $T_B$. The thermal resistance of a rod is given by $R = \frac{L}{kA}$. Since all rods have the same material $(k)$ and same cross-section $(A)$,the thermal resistance is proportional to the length $(R \propto L)$.
Let $R_{AB}, R_{BC}, R_{BD}$ be the resistances of rods $AB, BC, BD$ respectively.
Given that there is no heat flow in $AB$,the temperature at $B$ must be equal to the temperature at $A$. Therefore,$T_B = T_A = 20^{\circ} C$.
Since there is no heat flow in $AB$,all the heat flowing from $D$ to $B$ must flow from $B$ to $C$. Thus,the heat current $H_{DB} = H_{BC}$.
Using the formula $H = \frac{\Delta T}{R}$,we have $\frac{T_D - T_B}{R_{BD}} = \frac{T_B - T_C}{R_{BC}}$.
Substituting the values: $\frac{90 - 20}{R_{BD}} = \frac{20 - 0}{R_{BC}}$.
$\frac{70}{R_{BD}} = \frac{20}{R_{BC}}$.
$\frac{R_{BD}}{R_{BC}} = \frac{70}{20} = \frac{7}{2}$.
Since $R \propto L$,we have $\frac{L_{BD}}{L_{BC}} = \frac{7}{2}$.

(C) Let the temperature of the copper-iron junction be $\theta$.
Since the rods are joined in series and the surfaces are thermally insulated,the rate of heat flow through both rods must be the same in the steady state.
Let $K_1 = 386.4 \ W/m-K$ (copper),$l_1 = 0.75 \ m$,$T_1 = 100^{\circ} C$.
Let $K_2 = 48.46 \ W/m-K$ (iron),$l_2 = 1.25 \ m$,$T_2 = 0^{\circ} C$.
The cross-sectional area $A$ is the same for both.
The rate of heat flow is given by $H = \frac{KA(T_{high} - T_{low})}{l}$.
Equating the heat flow rates: $\frac{K_1 A (T_1 - \theta)}{l_1} = \frac{K_2 A (\theta - T_2)}{l_2}$.
$\frac{386.4 (100 - \theta)}{0.75} = \frac{48.46 (\theta - 0)}{1.25}$.
$515.2 (100 - \theta) = 38.768 \theta$.
$51520 - 515.2 \theta = 38.768 \theta$.
$553.968 \theta = 51520$.
$\theta = \frac{51520}{553.968} \approx 93^{\circ} C$.

(C) In a steady state,the rate of heat flow $(H)$ through both slabs connected in series is the same.
$H = \frac{k_1 A (\Delta T_A)}{L} = \frac{k_2 A (\Delta T_B)}{L}$
Since the thickness $(L)$ and cross-sectional area $(A)$ are the same for both slabs,we have:
$k_1 (\Delta T_A) = k_2 (\Delta T_B)$
Given $k_1 = \frac{k_2}{2}$,we can write $k_2 = 2k_1$.
Substituting this into the equation:
$k_1 (\Delta T_A) = 2k_1 (\Delta T_B)$
$\Delta T_A = 2 \Delta T_B$
We know the total temperature difference is $\Delta T_A + \Delta T_B = 12^{\circ} C$.
Substituting $\Delta T_A = 2 \Delta T_B$ into the total temperature equation:
$2 \Delta T_B + \Delta T_B = 12^{\circ} C$
$3 \Delta T_B = 12^{\circ} C \implies \Delta T_B = 4^{\circ} C$
Therefore,the temperature difference across slab $A$ is:
$\Delta T_A = 12^{\circ} C - 4^{\circ} C = 8^{\circ} C$.

(A) Let the thickness of slab $B$ be $d$ and its thermal conductivity be $K$. Then,for slab $A$,the thickness is $2d$ and thermal conductivity is $2K$.
Let the temperature at the contact surface be $T$.
Since the slabs are in series,the rate of heat flow $H$ through both slabs must be the same:
$H = \frac{K_A A (T_1 - T)}{d_A} = \frac{K_B A (T - T_2)}{d_B}$
Substituting the given values:
$\frac{2K \cdot A (100 - T)}{2d} = \frac{K \cdot A (T - 25)}{d}$
$(100 - T) = (T - 25)$
$2T = 125$
$T = 62.5^{\circ} C$

(B) Let the temperature at junction $B$ be $\theta$.
Since the rods $BC$ and $BD$ are connected to the same temperature source $(80^{\circ} C)$ at their outer ends,they act as parallel thermal resistances.
Let $R$ be the thermal resistance of each rod.
The equivalent resistance of the parallel combination of rods $BC$ and $BD$ is $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$.
At steady state,the heat current flowing through rod $AB$ must be equal to the sum of heat currents flowing through rods $BC$ and $BD$.
Using the formula for heat current $\frac{Q}{t} = \frac{\Delta T}{R}$,we have:
$\frac{\theta - 20}{R} = \frac{80 - \theta}{R/2}$
$\theta - 20 = 2(80 - \theta)$
$\theta - 20 = 160 - 2\theta$
$3\theta = 180$
$\theta = 60^{\circ} C$
Thus,the temperature at junction $B$ is $60^{\circ} C$.

(A) When two rods are connected in series,the total thermal resistance $R_{eq}$ is the sum of individual thermal resistances $R_1$ and $R_2$.
The thermal resistance of a rod is given by $R = \frac{l}{kA}$.
For the two rods in series: $R_{eq} = R_1 + R_2$.
$\frac{l_1+l_2}{k A} = \frac{l_1}{k_1 A} + \frac{l_2}{k_2 A}$.
Canceling the area $A$ from both sides:
$\frac{l_1+l_2}{k} = \frac{l_1}{k_1} + \frac{l_2}{k_2}$.
$\frac{l_1+l_2}{k} = \frac{k_2 l_1 + k_1 l_2}{k_1 k_2}$.
Therefore,$k = \frac{(l_1+l_2) k_1 k_2}{k_2 l_1 + k_1 l_2}$.

(A) For two rods of equal length $\ell$ and cross-sectional area $A$ joined in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances $R_1$ and $R_2$.
$R_{eq} = R_1 + R_2$
Since $R = \frac{\ell}{KA}$,we have:
$\frac{2\ell}{K_{eff} A} = \frac{\ell}{K_1 A} + \frac{\ell}{K_2 A}$
Canceling $\frac{\ell}{A}$ from both sides:
$\frac{2}{K_{eff}} = \frac{1}{K_1} + \frac{1}{K_2}$
Given $K_1 = 3$ and $K_2 = 4$:
$\frac{2}{K_{eff}} = \frac{1}{3} + \frac{1}{4} = \frac{4+3}{12} = \frac{7}{12}$
$K_{eff} = \frac{24}{7} \approx 3.43$ units.

(A) Let the thermal resistance of rod $x$ be $R = \frac{\ell}{k_x A}$. Since the thermal conductivity of $x$ is $3$ times that of $y$ $(k_x = 3k_y)$,the thermal resistance of rod $y$ is $3R$.
From the figure,the network between $B$ and $E$ consists of two parallel branches: one with rods $y$ and $y$ (total resistance $3R + 3R = 6R$) and one with rods $x$ and $x$ (total resistance $R + R = 2R$).
The equivalent resistance $R_{BE}$ between $B$ and $E$ is $\frac{1}{R_{BE}} = \frac{1}{6R} + \frac{1}{2R} = \frac{1+3}{6R} = \frac{4}{6R} = \frac{2}{3R}$,so $R_{BE} = 1.5R$.
The total resistance of the circuit is $R_{total} = R_{AB} + R_{BE} + R_{EF} = R + 1.5R + 3R = 5.5R = \frac{11R}{2}$.
The total heat current $H = \frac{100 - 40}{5.5R} = \frac{60}{5.5R} = \frac{120}{11R}$.
Now,$T_B = 100 - H \cdot R = 100 - \frac{120}{11R} \cdot R = 100 - 10.91 \approx 89.09^{\circ}C$.
And $T_E = 40 + H \cdot (3R) = 40 + \frac{120}{11R} \cdot 3R = 40 + 32.73 \approx 72.73^{\circ}C$.
Thus,$T_B \approx 89^{\circ}C$ and $T_E \approx 73^{\circ}C$.