Efficiency of Carnot engine is 50% and temperature of sink is 500, K . If temperature of source is k

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11.Thermodynamics

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The efficiency of Carnot engine is $50\%$ and temperature of sink is $500\, K$. If the temperature of source is kept constant and its efficiency is to be raised to $60\%$, then the required temperature of the sink will be........ $K$

A

$600$

B

$500$

C

$400$

D

$100$

Solution

$1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{50}{100}$

$\frac{500}{\mathrm{T}_{1}}=\frac{1}{2}$

$\Rightarrow \mathrm{T}_{1}=1000 \mathrm{K}$

$1-\frac{\mathrm{T}_{2}}{1000}=\frac{60}{100}$

$\frac{\mathrm{T}_{2}}{1000}=\frac{2}{5}$

$\mathrm{T}_{2}=400 \mathrm{K}$

Standard 11

Physics

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