Question 731: Thermodynamics & KTG

Question

Two rods of same length \& material transfer a given amount of heat in 12 s when they are joined end to end. But when they are joined length wise parallel to each other they will transfer same amount of heat in same condition in time

Accepted Answer

Let the heat transferred be Q.
When rods are joined end to end, heat transferred by each rod
$$=\mathrm{Q}=\frac{\mathrm{KA} 	riangle 	heta}{2} 	imes 12\quad	ext{... (i)}$$
When rods are joined lengthwise,
$$\mathrm{Q}=\frac{\mathrm{K} 2 \mathrm{~A} \Delta 	heta}{l} \mathrm{t}\quad	ext{... (ii)}$$
From equation (i) and (ii),
$$\begin{aligned}
& \frac{\mathrm{K} 2 \mathrm{~A} \Delta 	heta}{l} \mathrm{t}=\frac{\mathrm{KA} \Delta 	heta}{2 l} 	imes 12 \
	herefore \quad & \mathrm{t}=\frac{12}{2 	imes 2}=3 \mathrm{~s}
\end{aligned}$$

Answer

24 s

Answer

3 s

Answer

1.5 s

Answer

48 s

Thermodynamics & KTG

Two rods of same length \& material transfer a given amount of heat in 12 s when they are joined end to end. But when they are joined length wise parallel to each other they will transfer same amount of heat in same condition in time

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