Two plates A and B have thermal conductivities 84 ~Wm^-1 ~K^-1 and 126 ~Wm^-1 ~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 A and B are kept at 100^\circ C and 0 ^\circ C respectively, then the temperature of the surface of contact in steady state is \underline\hspace2cm ^\circ C .

Heat and Thermodynamics PYQ — JEE Main 2023 · 13 Apr · Shift 2 · Q66 with Solution

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Correct answer: 40

Set up steady-state heat conduction

Since the two plates are in contact, have the same thickness and same area, and heat flows steadily through them, the heat current through both plates must be the same.

Let the contact temperature be $T$ .

Plate $A$ : thermal conductivity $k_A = 84\,\text{W m}^{-1}\text{K}^{-1}$
Plate $B$ : thermal conductivity $k_B = 126\,\text{W m}^{-1}\text{K}^{-1}$

Outer temperatures are:

Left surface of $A$ : $100^\circ C$
Right surface of $B$ : $0^\circ C$

Apply Fourier's law

For plate $A$ ,

Q = \frac{k_A A (100 - T)}{L}

For plate $B$ ,

Q = \frac{k_B A (T - 0)}{L}

In steady state, these are equal:

\frac{84A(100-T)}{L} = \frac{126A(T)}{L}

Cancel $A$ and $L$ :

84(100-T)=126T

Solve for $T$

8400 - 84T = 126T

8400 = 210T

T = \frac{8400}{210} = 40

Final answer

The steady-state temperature of the contact surface is

\boxed{40^\circ C}

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