A Carnot engine whose sink is at 300 K has an efficiency of 40 % . By how much should the temperature of source be increased so as to increase its efficiency by 50 % of original efficiency.

Question

A Carnot engine whose sink is at 300 K has an efficiency of 40 % . By how much should the temperature of source be increased so as to increase its efficiency by 50 % of original efficiency. (A) 275K (B) 325K (C) 250K (D) 380K

Tardigrade Admin · Accepted Answer

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied, i.e.,

η = \frac{Work done}{Heat supplied}

= \frac{W}{Q _{1}} = \frac{Q _{1} - Q _{2}}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{T _{2}}{T _{1}}

Here,

T_{1}

is the temperature of source and

T_{2}

is the temperature of sink.
As given,

η = 40% = \frac{40}{100} = 0.4

and

T_{2} = 300 K

So,

0.4 = 1 - \frac{300}{T _{1}}

.

\Rightarrow T_{1} = \frac{300}{1 - 0.4} = \frac{300}{0.6} = 500 K

Let, temperature of the source be increased by

x K,

then efficiency becomes

η^{^{'}} = 40% + 50%

of

η

= \frac{40}{100} + \frac{50}{100} \times 0.4 = 0.4 + 0.5 \times 0.4 = 0.6

Hence,

0.6 = 1 - \frac{300}{500 + x}

\Rightarrow \frac{300}{500 + x} = 0.4

\Rightarrow 500 + x = \frac{300}{0.4} = 750

x = 750 - 500 = 250 K

Q. A Carnot engine whose sink is at $300 K$ has an efficiency of $40%$ . By how much should the temperature of source be increased so as to increase its efficiency by $50%$ of original efficiency.

$275 K$

$325 K$

$250 K$

$380 K$

Q. A Carnot engine whose sink is at 300K has an efficiency of 40% . By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency.

275K

325K

250K

380K

Q. A Carnot engine whose sink is at $300 K$ has an efficiency of $40%$ . By how much should the temperature of source be increased so as to increase its efficiency by $50%$ of original efficiency.

$275 K$

$325 K$

$250 K$

$380 K$