Question 512: Thermodynamics & KTG

Question

Three rods of Copper, Brass and Steel are welded together to form a $$Y$$ shaped structure. Area of cross - section of each rod $$ = 4c{m^2}.$$ End of copper rod is maintained at $${100^ \circ }C$$ where as ends of brass and steel are kept at $${0^ \circ }C$$. Lengths of the copper, brass and steel rods are $$46,$$ $$13$$ and $$12$$ $$cms$$ respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are $$0.92, 0.26$$ and $$0.12$$ $$CGS$$ units respectively. Rate of heat flow through copper rod is:

Accepted Answer

Rate of heat flow is given by,
$$Q = {{KA\left( {{	heta _1} - {	heta _2}} ight)} \over l}$$
Where, $$K=$$ coefficient of thermal conductivity $$l=$$ length of rod and $$A=$$ Area of cross-section of rod
If the junction temperature is $$T,$$ then
$${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$$
$${{0.92 	imes 4\left( {100 - T} ight)} \over {46}} = {{0.26 	imes 4 	imes \left( {T - 0} ight)} \over {13}} + {{0.12 	imes 4 	imes \left( {T - 0} ight)} \over {12}}$$
$$ \Rightarrow 200 - 2T = 2T + T$$
$$ \Rightarrow T = {40^ \circ }C$$
$$	herefore$$ $${Q_{Copper}} = {{0.92 	imes 4 	imes 60} \over {46}} = 4.8\,cal/s$$

Answer

$$1.2$$ $$cal/s$$

Answer

$$2.4$$ $$cal/s$$

Answer

$$4.8$$ $$cal/s$$

Answer

$$6.0$$ $$cal/s$$

Thermodynamics & KTG

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