KCET · Physics · Rotational Motion
What is the source temperature of the Carnot engine required to get \( 70 \% \) efficiency ? Given
sink temperature \( =27^{\circ} \mathrm{C} \)
- A \( 1000^{\circ} \mathrm{C} \)
- B \( 90^{\circ} \mathrm{C} \)
- C \( 270^{\circ} \mathrm{C} \)
- D \( 727^{\circ} \mathrm{C} \)
Answer & Solution
Correct Answer
(D) \( 727^{\circ} \mathrm{C} \)
Step-by-step Solution
Detailed explanation
Given, efficiency, \(\eta=70 \%\)
sink temperature, \(T_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}\)
Efficiency, \(\eta=\left(1-\frac{T_{2}}{T_{1}}\right) \times 100 \%\)
\(\Rightarrow 70=\left(1-\frac{300}{T_{1}}\right) \times 100\)
\(\Rightarrow \frac{70}{100}=1-\frac{300}{T_{1}}\)
\(\Rightarrow 0.7=1-\frac{300}{T_{1}}\)
\(\Rightarrow \frac{300}{T_{1}}=1-0.7\)
\(\Rightarrow \frac{300}{T_{1}}=0.3\)
\(T_{1}=\frac{300}{0.3}=1000 \mathrm{~K}\)
or \(T_{1}=(1000-273)^{\circ} \mathrm{C}=727^{\circ} \mathrm{C}\)
Therefore, source temperature \(=727^{\circ} \mathrm{C}\)
sink temperature, \(T_{2}=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}\)
Efficiency, \(\eta=\left(1-\frac{T_{2}}{T_{1}}\right) \times 100 \%\)
\(\Rightarrow 70=\left(1-\frac{300}{T_{1}}\right) \times 100\)
\(\Rightarrow \frac{70}{100}=1-\frac{300}{T_{1}}\)
\(\Rightarrow 0.7=1-\frac{300}{T_{1}}\)
\(\Rightarrow \frac{300}{T_{1}}=1-0.7\)
\(\Rightarrow \frac{300}{T_{1}}=0.3\)
\(T_{1}=\frac{300}{0.3}=1000 \mathrm{~K}\)
or \(T_{1}=(1000-273)^{\circ} \mathrm{C}=727^{\circ} \mathrm{C}\)
Therefore, source temperature \(=727^{\circ} \mathrm{C}\)
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