Question 870: Thermodynamics & KTG

Question

A Carnot engine with efficiency $$50 \%$$ takes heat from a source at $$600 \mathrm{~K}$$. To increase the efficiency to $$70 \%$$, keeping the temperature of the sink same, the new temperature of the source will be

Accepted Answer

$$\begin{aligned}
& T_H=600 \mathrm{k} \
& \eta=1-\frac{T_C}{T_H} \
& 	ext { but, } \eta=\frac{1}{2} \quad ....(	ext{Given} : \eta=50 \%)\
& \Rightarrow \frac{1}{2}=1-\frac{T_C}{600} \
& 	herefore \quad T_C=300 K
\end{aligned}$$
With $$\mathrm{T}_{\mathrm{C}}=300 \mathrm{~K}$$, the efficiency is increased to $$70 \%$$
$$	herefore \quad$$ New temperature of the source will be $$\mathrm{T}_{\mathrm{H}_{	ext {new }}}$$
$$\begin{array}{ll}
	herefore \quad \eta=1-\frac{300}{\mathrm{~T}_{\mathrm{H}_{	ext {new }}}} \
 \frac{300}{\mathrm{~T}_{\mathrm{H}_{	ext {new }}}}=1-\frac{7}{10} \quad \ldots .(\because \eta=70 \%) \
	herefore \quad \mathrm{T}_{\mathrm{H}_{	ext {new }}}=\frac{3000}{3}=1000 \mathrm{~K}
\end{array}$$

Answer

$$360 \mathrm{~K}$$

Answer

$$1000 \mathrm{~K}$$

Answer

$$900 \mathrm{~K}$$

Answer

$$300 \mathrm{~K}$$

Thermodynamics & KTG

A Carnot engine with efficiency $50 \% takes heat from a source at 600 \mathrm{~K}. To increase the efficiency to 70 \%$, keeping the temperature of the sink same, the new temperature of the source will be

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