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JEE Advanced 2011 · Thermal Properties of Matter

A composite block is made of slabs AA, BB, CC, DD and EE of different thermal conductivities (given in terms of a constant KK) and sizes (given in terms of length, LL) as shown in the figure. All slabs are of same width. Heat QQ flows only from left to right through the blocks. Then in steady state

Diagram for this Thermal Properties of Matter question (JEE Advanced 2011).

Answers: A. heat flow through A and E slabs are same. · C. temperature difference across slab E is smallest. · D. heat flow through C=C = heat flow through B + heat flow through D.

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Written solution

Step 1: Identify the slab connections.

Slabs B, C, and D are in parallel, while A and E are in series with this combination.

Step 2: Determine the heat flow rates.

The heat flow rate through series components is equal, while parallel heat flows divide by conductivity.

QA=QE=QB+QC+QDQ_A = Q_E = Q_B + Q_C + Q_D QB:QC:QD=3:4:5Q_B : Q_C : Q_D = 3 : 4 : 5 QC≠QB+QDQ_C \neq Q_B + Q_D

Step 3: Determine the temperature differences.

The temperature difference across each section depends on its thermal resistance.

R=lkAR = \frac{l}{kA} ΔT=QR\Delta T = Q R ΔTA:ΔTmid:ΔTE=0.30:0.60:0.10\Delta T_A : \Delta T_{\text{mid}} : \Delta T_E = 0.30 : 0.60 : 0.10 ΔTE=Q3kE\Delta T_E = \frac{Q}{3k_E}

Answer: In steady state, the heat flow rate through A is equal to that through E (QA=QEQ_A = Q_E), and the temperature difference across E is minimum.

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