Question

On doubling the temperature of source, the efficiency of a
heat engine becomes triple. The new efficiency is :-

​

Answer 1

If we increase the temperature of a source, it’s efficiency will increase

Answer 2

The new efficiency of the given heat engine is 0.6 0r 60%.

Given,

The initial temperature of the source, T$_1$

The initial temperature of the sink, T$_2$

Initial efficiency η$_i$ and final efficiency η$_f$

On doubling T$_1$, η$_f$ = 2η$_i$

To find,

η$_f$, the new efficiency

Solution,

This question can be solved easily with the formula for efficiency.

We know,

η$_i$ = 1 - T$_2$/T$_1$  

η$_f$ = 3η$_i$ = 1 - T$_2$/2T$_1$  _____ (1)

⇒ 3 (1 - T$_2$/T$_1$) =  1 - T$_2$/2T$_1$

⇒ 3 - 3T$_2$/T$_1$ =  1 - T$_2$/2T$_1$

⇒ 2 = 3T$_2$/T$_1$ - T$_2$/2T$_1$

⇒ 2 = (6T$_2$ - T$_2$)/2T$_1$

⇒ 2 = 5T$_2$/2T$_1$

⇒ T$_2$/T$_1$  = 4/5 ____ (2)

From equation (1) and (2) we get,

η$_f$ = 1 - 1/2 × 4/5

⇒ η$_f$ = 0.6 = 60%

∴ The new efficiency η$_f$ is 0.6 or 60%.