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The increase in entropy in $\mathrm{JK}^{-1}$ of a substance when it absorbs $1 \mathrm{~kJ}$ of heat energy at $3 \mathrm{~K}$ is
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1658 Upvotes
Verified Answer
The correct answer is:
333.3
Given,
$$
\begin{aligned}
& \text { Heat }(q)=1 \mathrm{~kJ}=1000 \mathrm{~J} \\
& \text { Temperature }=3 \mathrm{~K} \\
& \text { Entropy } \begin{aligned}
(\Delta S) & =\frac{\text { Absorb heat }(q)}{\text { Temperature }(T)} \\
& =\frac{1000 \mathrm{~J}}{3 \mathrm{~K}}
\end{aligned}
\end{aligned}
$$
Entropy $(S)=333.3 \mathrm{~J} / \mathrm{K}$
$$
\begin{aligned}
& \text { Heat }(q)=1 \mathrm{~kJ}=1000 \mathrm{~J} \\
& \text { Temperature }=3 \mathrm{~K} \\
& \text { Entropy } \begin{aligned}
(\Delta S) & =\frac{\text { Absorb heat }(q)}{\text { Temperature }(T)} \\
& =\frac{1000 \mathrm{~J}}{3 \mathrm{~K}}
\end{aligned}
\end{aligned}
$$
Entropy $(S)=333.3 \mathrm{~J} / \mathrm{K}$
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