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Two rods, one made of copper and the other made of steel, of the same length and same cross-sectional area are joined together. The thermal conductivity of copper and steel are 385 J s−1K−1m−1385\text{ J s}^{-1}\text{K}^{-1}\text{m}^{-1} and 50 J s−1K−1m−150\text{ J s}^{-1}\text{K}^{-1}\text{m}^{-1} respectively. The free ends of copper and steel are held at 100∘C100^{\circ}\text{C} and 0∘C0^{\circ}\text{C} respectively. The temperature at the junction is, nearly:

A

12∘C12^{\circ}\text{C}

B

50∘C50^{\circ}\text{C}

C

73∘C73^{\circ}\text{C}

D

88.5∘C88.5^{\circ}\text{C}

Step-by-Step Solution

Let the temperature of the junction be TT. In the steady state, the rate of heat flow through the copper rod is equal to the rate of heat flow through the steel rod. dQdt=KcA(100−T)L=KsA(T−0)L\frac{dQ}{dt} = \frac{K_c A (100 - T)}{L} = \frac{K_s A (T - 0)}{L} Since the length (LL) and cross-sectional area (AA) of both rods are the same, they cancel out: Kc(100−T)=Ks(T−0)K_c(100 - T) = K_s(T - 0) Given Kc=385 J s−1K−1m−1K_c = 385\text{ J s}^{-1}\text{K}^{-1}\text{m}^{-1} and Ks=50 J s−1K−1m−1K_s = 50\text{ J s}^{-1}\text{K}^{-1}\text{m}^{-1}: 385(100−T)=50T385(100 - T) = 50T 38500−385T=50T38500 - 385T = 50T 435T=38500435T = 38500 T=38500435≈88.5∘CT = \frac{38500}{435} \approx 88.5^{\circ}\text{C}

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