JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 23)
Two plates $$\mathrm{A}$$ and $$\mathrm{B}$$ have thermal conductivities $$84 ~\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$$ and $$126 ~\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $$\mathrm{A}$$ and $$\mathrm{B}$$ are kept at $$100^{\circ} \mathrm{C}$$ and $$0{ }^{\circ} \mathrm{C}$$ respectively, then the temperature of the surface of contact in steady state is _____________ $${ }^{\circ} \mathrm{C}$$.
Answer
40
Explanation
Let's denote the temperature at the surface of contact as T. We can find this temperature by considering the heat transfer through each plate when the system reaches steady state. At steady state, the rate of heat transfer through both plates A and B is the same.
We can use the formula for heat transfer rate through a plate:
$$Q = kA\frac{T_2 - T_1}{d}$$
Where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures on either side of the plate, and d is the thickness of the plate.
For plate A:
$$Q_A = k_A A\frac{T_A - T}{d}$$
For plate B:
$$Q_B = k_B A\frac{T - T_B}{d}$$
Since the heat transfer rate is the same through both plates in steady state:
$$Q_A = Q_B$$
We can now substitute the given values for thermal conductivities and temperatures:
$$84A\frac{100 - T}{d} = 126A\frac{T - 0}{d}$$
Notice that the surface area (A) and thickness (d) are the same for both plates, so they cancel out:
$$84(100 - T) = 126T$$
Now, we can solve for T:
$$8400 - 84T = 126T$$
$$210T = 8400$$
$$T = \frac{8400}{210}$$
$$T = 40$$
So, the temperature of the surface of contact in steady state is $$40{ }^{\circ} \mathrm{C}$$.
We can use the formula for heat transfer rate through a plate:
$$Q = kA\frac{T_2 - T_1}{d}$$
Where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures on either side of the plate, and d is the thickness of the plate.
For plate A:
$$Q_A = k_A A\frac{T_A - T}{d}$$
For plate B:
$$Q_B = k_B A\frac{T - T_B}{d}$$
Since the heat transfer rate is the same through both plates in steady state:
$$Q_A = Q_B$$
We can now substitute the given values for thermal conductivities and temperatures:
$$84A\frac{100 - T}{d} = 126A\frac{T - 0}{d}$$
Notice that the surface area (A) and thickness (d) are the same for both plates, so they cancel out:
$$84(100 - T) = 126T$$
Now, we can solve for T:
$$8400 - 84T = 126T$$
$$210T = 8400$$
$$T = \frac{8400}{210}$$
$$T = 40$$
So, the temperature of the surface of contact in steady state is $$40{ }^{\circ} \mathrm{C}$$.
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