Two identiacal metal blocks $L$ and $M$ (specific heat $= 0.4 k J / k g K$ ), each having a mass of $5 k g$ , are initialy at $313 K$ . A reversible refrigerator extracts heat from block $L$ and rejects heat to block $M$ untill the tmperature of block $L$ reaches $293 K$ . The final temperature in (K) of block $M$ is
334.36

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Solution

The correct option is A 334.36

Given data:
Specified heath,
$c = 0.4 k J k g K$ each body
Mass: $m = 5, k g$ eachbody
Initial temperature,
$T_{1} = 313 K$
Final temperature of block $L$
$T_{2} = 293 K$
Let $T_{1} =$ Final temperature of block $M$
Entroy change of block $L$ ,
$Δ S_{L} = m c l o g_{e} \frac{T_{2}}{T_{1}}$
Entropy change of block $M$ ,
$Δ S_{M} = m c l o g_{e} \frac{T_{f}}{T_{1}}$
Entropy change of universe,
$Δ S_{u n i v} = Δ S_{L} + Δ S_{M}$
$= m c l o g_{e} \frac{T_{2}}{T_{1}} + m c l o g_{e} \frac{T_{f}}{T_{1}}$
For a reversible refrigeration,
$Δ S_{u n i v} = 0$
$m c l o g_{e} \frac{T_{2}}{T_{1}} + m c l o g_{e} \frac{T_{f}}{T_{1}} = 0$
$m c l o g_{e} \frac{T_{2} T_{f}}{T_{1}^{2}} = 0$
or $l o g_{e} \frac{T_{2} T_{f}}{T_{1}^{2}} = 0$
or $\frac{T_{2} T_{f}}{T_{1}^{2}} == e^{o} = 1$
$T_{f} = \frac{T_{1}^{2}}{T_{2}} = \frac{(313)^{2}}{(293)} = 334.36 K$

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