Question

A Carnot engine with efficiency 40% takes heat from a source at 500 K. To increase the efficiency to 60%, keeping temperature of the sink same, the new temperature of the source will be ______.

विकल्प

• 900 K

• 750 K

• 1000 K

• 450 K

Answer 1

A Carnot engine with efficiency 40% takes heat from a source at 500 K. To increase the efficiency to 60%, keeping temperature of the sink same, the new temperature of the source will be 750 K.

Explanation:

\[T_H=500K\]

\[\eta=1-\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{T}_{\mathrm{H}}}\]

but, \[\eta=\frac{2}{5}\]         ....(Given: \[\eta\] = 40%)

\[\Rightarrow\quad\frac{2}{5}=1-\frac{\mathrm{T_C}}{500}\]

\[\therefore\] \[T_C=300 K\]

With \[T_C=300 K\], the efficiency is increased to 60%

\[\therefore\] New temperature of the source willl be \[\mathrm{T}_{\mathrm{H}_{\mathrm{new}}}\]

\[\therefore\quad\eta=1-\frac{300}{\mathrm{T}_{\mathrm{H_{new}}}}\]

\[\frac{300}{\mathrm{T}_{\mathrm{H}_{\mathrm{new}}}}=1-\frac{3}{5}\quad....(\because\eta=60\%)\]

\[\therefore\quad\mathrm{T}_{\mathrm{H_{new}}}=\frac{5}{2}\times300=750\mathrm{K}\]