A brass rod of length 2 m and radius 1 cm h | vedclass

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(C) Given: Rate of heat flow $\frac{Q}{t} = 0.5 \ cal \ s^{-1}$, Radius $r = 1 \ cm$, Length $L = 2 \ m = 200 \ cm$, Temperature $T_1 = 250 \ ^oC$, Th

Vedclass · Accepted Answer

$127.6$

Vedclass · Answer

(C) Given: Rate of heat flow $\frac{Q}{t} = 0.5 \ cal \ s^{-1}$, Radius $r = 1 \ cm$, Length $L = 2 \ m = 200 \ cm$, Temperature $T_1 = 250 \ ^oC$, Thermal conductivity $K = 0.26 	ext{ cal s}^{-1} 	ext{ cm}^{-1} 	ext{ } ^\circ	ext{C}^{-1}$.
Area of cross-section $A = \pi r^2 = 3.142 	imes (1)^2 = 3.142 \ cm^2$.
The formula for steady-state heat conduction is $\frac{Q}{t} = \frac{KA(T_1 - T_2)}{L}$.
Rearranging for temperature difference: $(T_1 - T_2) = \frac{(Q/t) 	imes L}{KA}$.
Substituting the values: $(T_1 - T_2) = \frac{0.5 	imes 200}{0.26 	imes 3.142} = \frac{100}{0.81692} \approx 122.4 \ ^oC$.
Therefore, the temperature of the other end $T_2 = T_1 - 122.4 \ ^oC = 250 \ ^oC - 122.4 \ ^oC = 127.6 \ ^oC$.

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