Logo
Jeeify
MCQEasyJEE 2023Heat Transfer (Conduction, Convection, Radiation)

JEE Physics 2023 Question with Solution

Two plates AA and BB have thermal conductivities 84 W/m\cdotK84 \, \text{W/m\cdot K} and 126 W/m\cdotK126 \, \text{W/m\cdot K} respectively. They have the same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of AA and BB are kept at 100∘C100^\circ \text{C} and 0∘C0^\circ \text{C} respectively, then the temperature of the surface of contact in steady state is _____ ∘C^\circ \text{C}.

  • A

    40∘C40^\circ \text{C}

  • B

    50∘C50^\circ \text{C}

  • C

    60∘C60^\circ \text{C}

  • D

    70∘C70^\circ \text{C}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Two plates have thermal conductivities KA=84 W/m\cdotKK_A = 84 \, \text{W/m\cdot K} and KB=126 W/m\cdotKK_B = 126 \, \text{W/m\cdot K}. They have the same surface area and the same thickness. The outer surface temperatures are TA=100∘CT_A = 100^\circ \text{C} and TB=0∘CT_B = 0^\circ \text{C}.

Find: The steady-state temperature TT of the contact surface.

In steady state, the heat current through plate AA equals the heat current through plate BB.

KAA(TA−T)L=KBA(T−TB)LK_A A \frac{(T_A - T)}{L} = K_B A \frac{(T - T_B)}{L}

Substituting the given values,

84⋅(100−T)=126⋅(T−0)84 \cdot (100 - T) = 126 \cdot (T - 0)

Simplifying,

8400−84T=126T8400 - 84T = 126T

Rearranging,

8400=210T8400 = 210T

Therefore,

T=8400210=40∘CT = \frac{8400}{210} = 40^\circ \text{C}

Therefore, the temperature of the contact surface is 40∘C40^\circ \text{C}. The correct option is A.

Common mistakes

  • Using the conductivities in the wrong order. Since plate AA is on the 100∘C100^\circ \text{C} side and plate BB is on the 0∘C0^\circ \text{C} side, the temperature drops must be assigned correctly. Write the steady-state heat currents for each plate before substituting values.

  • Assuming the contact temperature is the simple average of 100∘C100^\circ \text{C} and 0∘C0^\circ \text{C}. This is wrong because the temperature division depends on thermal conductivities, not only on endpoint temperatures. Use equality of heat current instead.

Practice more Heat Transfer (Conduction, Convection, Radiation) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Heat Transfer (Conduction, Convection, Radiation)

Related questions