MHT CET · Physics · Thermal Properties of Matter
The rate of flow of heat through a metal rod with temperature difference \(40^{\circ} \mathrm{C}\) is \(1600 \mathrm{cal} / \mathrm{s}\). The thermal resistance of metal rod in \({ }^{\circ} \mathrm{C} \mathrm{s} / \mathrm{cal}\) is
- A \(0.025\)
- B \(0.25\)
- C \(2.5\)
- D \(40\)
Answer & Solution
Correct Answer
(A) \(0.025\)
Step-by-step Solution
Detailed explanation
Given: rate of flow of heat (conduction rate)
\(
\mathrm{P}_{\text {cond }}=1600 \mathrm{cal} / \mathrm{s}
\)
Thermal resistance,
\(
\begin{aligned}
& \mathrm{R}_{\mathrm{T}}=\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}} \\
& \mathrm{R}_{\mathrm{T}}=\frac{40}{1600} \\
& \mathrm{R}_{\mathrm{T}}=0.025^{\circ} \mathrm{Cs} / \mathrm{cal}
\end{aligned}
\)
\(
\mathrm{P}_{\text {cond }}=1600 \mathrm{cal} / \mathrm{s}
\)
Thermal resistance,
\(
\begin{aligned}
& \mathrm{R}_{\mathrm{T}}=\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}} \\
& \mathrm{R}_{\mathrm{T}}=\frac{40}{1600} \\
& \mathrm{R}_{\mathrm{T}}=0.025^{\circ} \mathrm{Cs} / \mathrm{cal}
\end{aligned}
\)
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