Question

Consider a Carnot cycle operating between source temperature 750 K and sink temperature 350 K producing 1.25 kJ of mechanical work per cycle, the heat transferred to the engine by the reservoirs

• A. 1.34 kJ
• B. 2.34 kJ
• C. 3.34 kJ
• D. 4.34 kJ

Answer 1

Answer: (B)

2.34 kJ

Answer 2

Here, $\mathrm { T } _ { 1 } = 750 \mathrm {~K} ; \mathrm { T } _ { 2 } = 350 \mathrm {~K}$

$\mathrm { W } = 1.25 \mathrm {~kJ} = 1250 \mathrm {~J}$

Now efficiency

$\eta = 1 - \frac { T _ { 2 } } { T _ { 1 } } = 1 - \frac { 350 } { 750 } = \frac { 400 } { 750 } = \frac { 8 } { 15 }$

As $\eta = \frac { \mathrm { W } } { \mathrm { Q } _ { 1 } }$

$\therefore Q _ { 1 } = \frac { W } { \eta } = \frac { 1250 } { 8 } \times 15 = 2343.75 \mathrm {~J} = 2.34 \mathrm {~kJ}$