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Calculate the entropy change \mathrm{(\Delta S)} of the surroundings when \mathrm{1000 \mathrm{~J}} of heat is transferred reversibly from a hot reservoir at \mathrm{400 \mathrm{~K}} to a cold reservoir at 300 K.

Option: 1

\mathrm{-2.50 \mathrm{~J} / \mathrm{K}}


Option: 2

\mathrm{0.4 \mathrm{~J} / \mathrm{K}}


Option: 3

\mathrm{-0.6 \mathrm{~J} / \mathrm{K}}


Option: 4

\mathrm{0.8 \mathrm{~J} / \mathrm{K}}


Answers (1)

best_answer

The entropy change of the surroundings is given by the formula:

\mathrm{ \ \Delta S_{\text {surroundings }}=-\frac{q}{T} }
Step 1: Convert temperatures to Kelvin

\mathrm{ \begin{aligned} & \mathrm{ST}_{\text {hot }}=400 \\\\ & \mathrm{~K}, \mathrm{~T}_{\text {cold }}=300 \mathrm{~K} \end{aligned} }

Step 2: Calculate \mathrm{\Delta S_{\text {surroundings }} }

\mathrm{ \Delta S_{\text {surroundings }}=-\frac{q}{T}=-\frac{1000 \mathrm{~J}}{400 \mathrm{~K}}=-2.5 \mathrm{~J} / \mathrm{K} }

Therefore, the entropy change \mathrm{(\Delta S)} of the surroundings is \mathrm{-2.5 \mathrm{~J} / \mathrm{K}}.

So, the correct option is (1)

Posted by

Ritika Jonwal

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