If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature $27^{\circ}…

If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature $27^{\circ} \mathrm{C}$, the temperature inside the refrigerator is
  1. $240 \mathrm{~K}$
  2. $250 \mathrm{~K}$
  3. $230 \mathrm{~K}$
  4. $260 \mathrm{~K}$

Solution

Here, Coefficient of performance $(\beta)=5$ $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=(27+273) \mathrm{K}=300 \mathrm{~K}$
As, $\beta=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}} \Rightarrow 5=\frac{\mathrm{T}_{2}}{300-\mathrm{T}_{2}}$
or $1500-5 \mathrm{~T}_{2}=\mathrm{T}_{2}$ or $6 \mathrm{~T}_{2}=1500$
$\therefore \mathrm{T}_{2}=\frac{1500}{6}=250 \mathrm{~K}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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