Question

As shown in the figure, six rods of same geometry are connected and maintained at temperatures \(100^\circ\text{C}\) and \(40^\circ\text{C}\). The temperature at points \(A\) and \(B\) are:

• A. \(T_A = 73^\circ\text{C},\; T_B = 89^\circ\text{C}\)
• B. \(T_A = 85^\circ\text{C},\; T_B = 75^\circ\text{C}\)
• C. \(T_A = 89^\circ\text{C},\; T_B = 73^\circ\text{C}\)
• D. \(T_A = 74^\circ\text{C},\; T_B = 88^\circ\text{C}\)

Hint:

For steady-state heat conduction problems:
Replace rods with thermal resistances
Use symmetry to simplify complex networks
Temperature drop is proportional to resistance

Answer 1

The Correct Option is C

Calculation: - **(I) Isothermal Reversible Work:** \( |w| = nRT \ln(V_2/V_1) = 1(8.314)(300) \ln(10) \). \( |w| \approx 2494 \times 2.303 \approx 5743\,\text{J} = 5.74\,\text{kJ} \). Matches **(D)**. - **(II) Isothermal Irreversible Work:** \( |w| = P_{ext} \Delta V = (3 \times 10^3\,\text{Pa}) (3-1\,\text{m}^3) = 6 \times 10^3\,\text{J} = 6\,\text{kJ} \). Matches **(B)**. - **(III) Enthalpy Change:** \( \Delta H = n C_p \Delta T = 1 (5R/2) (400) = 1000 R = 8314\,\text{J} \approx 8.32\,\text{kJ} \). *(Correction: Option A is 8.32. Match III-A).* - **(IV) Internal Energy Change:** \( \Delta U = n C_v \Delta T = 1 (3R/2) (320) = 480 R = 480(8.314) \approx 3990\,\text{J} \approx 4\,\text{kJ} \). Matches **(C)**. Sequence: I-D, II-B, III-A, IV-C. Option (2).