A Carnot engine whose low temperature reservoir is at 7,°C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By

Q: A Carnot engine whose low temperature reservoir is at $7\,°C$ has an efficiency of $50\%$. It is desired to increase the efficiency to $70\%$. By how many degrees should the temperature of the high temperature reservoir be increased ....... $K$

d (d) Initially $\eta = \frac{{{T_1} - {T_2}}}{{{T_1}}}$==> $0.5 = \frac{{{T_1} - (273 + 7)}}{{{T_1}}}$ ==> $\frac{1}{2} = \frac{{{T_1} - 280}}{{{T_1}}}$==> ${T_1} = 560K$ Finally ${\eta _1}' = \frac{{{T_1}'\, - {T_2}}}{{{T_1}^\prime }}$==>$0.7 = \frac{{{T_1}'\, - \,(273 + 7)}}==>{{{T_1}^\prime }}$==>${T_1}' = 933K$ $\therefore$ increase in temperature $ = 933 - 560 = 373K \approx 380K$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A Carnot engine whose low temperature reservoir is at $7\,°C$ has an efficiency of $50\%$. It is desired to increase the efficiency to $70\%$. By how many degrees should the temperature of the high temperature reservoir be increased ....... $K$

A$840$
B$280$
C$560$
D$380$

Medium

STD 11 Science
NEET
STD 11 - 11. thermodynamics
Physics

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