Question

A heat engine has an efficiency $\eta$. Temperatures of source and sink are each decreased by 100 K. The efficiency of the engine

• A. Increases
• B. Decreases
• C. Remains constant
• D. Becomes 1

Answer 1

Answer: (A)

Increases

Answer 2

$\eta = 1 - \frac { T _ { 2 } } { T _ { 1 } } = \frac { T _ { 1 } - T _ { 2 } } { T _ { 1 } }$

When and $\mathrm { T } _ { 2 }$ both are decreased by 100 K each, stays constant $\mathrm { T } _ { 1 }$ decreases.

$\therefore \eta$ increases