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EEJ MAIN Physics QUESTION #7068
Question 1
Three rods made of copper, brass, and steel are joined to form a Y-shape. Each rod has a cross-sectional area of $4 \text{ cm}^2$. The free end of the copper rod is at $100^\circ\text{C}$, while the ends of the brass and steel rods are at $0^\circ\text{C}$. Lengths are: Copper $= 46 \text{ cm}$, Brass $= 13 \text{ cm}$, Steel $= 12 \text{ cm}$. Thermal conductivities (CGS): Copper $= 0.92$, Brass $= 0.26$, Steel $= 0.12$. Find the heat flow rate 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 through the copper rod equals the sum of currents through brass and steel.

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

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

$0.02 (100 - T) = 0.02T + 0.01T \implies 2 - 0.02T = 0.03T \implies 5T = 200 \implies T = 40^\circ\text{C}$.

Rate of heat flow $H = \frac{0.92 \times 4 \times (100 - 40)}{46} = 0.02 \times 4 \times 60 = 4.8 \text{ cal/s}$.