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EEJ MAIN Physics QUESTION #7050
Question 1
A Y-shaped structure is created by welding together three rods of copper, brass, and steel. Each rod has a cross-sectional area of $4 \text{ cm}^2$. The copper rod's free end is at $100^\circ\text{C}$, while the ends of the brass and steel rods are at $0^\circ\text{C}$. The lengths of the copper, brass, and steel rods are $46 \text{ cm}$, $13 \text{ cm}$, and $12 \text{ cm}$, respectively. Given thermal conductivities (in CGS units) of $0.92$ for copper, $0.26$ for brass, and $0.12$ for steel, find the rate of heat flow through the copper rod.
  • $4.8 \text{ cal/s}$✔️
  • $6.0 \text{ cal/s}$
  • $1.2 \text{ cal/s}$
  • $2.4 \text{ cal/s}$
Correct Answer Explanation

Let the junction temperature be $T$. In steady state, the heat current entering the junction equals the heat current leaving it:

$H_{cu} = H_{br} + H_{st}$

$\frac{K_c A(100 - T)}{L_c} = \frac{K_b A(T - 0)}{L_b} + \frac{K_s A(T - 0)}{L_s}$

Substituting the values:

$\frac{0.92 \times 4 \times (100 - T)}{46} = \frac{0.26 \times 4 \times T}{13} + \frac{0.12 \times 4 \times T}{12}$

$0.02(100 - T) = 0.02T + 0.01T \implies 2 - 0.02T = 0.03T \implies T = 40^\circ\text{C}$

Rate of heat flow in copper rod $H_{cu} = 0.02 \times 4 \times (100 - 40) = 0.08 \times 60 = 4.8 \text{ cal/s}$.