Question

Solve this question please​

Answer 1

Solution : 250 K

Explanation:

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

n = Work done / Heat supplied

n = W / Q1 = Q1 - Q2 / Q1

n = 1 - Q2 - Q1 = 1 - T2 / T1

Here, T1 is the temperature of source, T2 is the temperature of sink, Q1 is the heat absorbed and Q2 heat rejected

As given, n = 40% = 40/100 = 0.4 and T2 = 300 K

So, 0.4 = 1 - 300/T1

=> T1 = 300 / 1 - 0.4 = 300 / 0.6 = 500 K

Let the temperature of the source be increased by x K, then efficiency becomes,

n' = 40% + 50% of n

n' = 40 / 100 + 50 / 100 × 0.4

n' = 0.4 + 0.5 × 0.4 = 0.6

Hence, 0.6 = 1 - 300 / 500 + x

=> 300 / 500 + x = 0.4

=> 500 + x = 300 / 0.4 = 750

∴ x = 750 - 500 = 250 K

∴ Option C is the correct answer !

Answer 2

$\large\boxed{\textsf{\textbf{\red{Question\::-}}}}$

• 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?

$\large\boxed{\textsf{\textbf{\red{Answer\::-}}}}$

• 250k

$\large\boxed{\textsf{\textbf{\red{Explanation\::-}}}}$

Temperature of sink

$T _{ L} = 300K$

Original efficiency

$η = 40\% = 0.4$

Let the initial temperature of the source be

$T_{ H}$

Using

$η = 1 - \frac{T_{ L } }{T _ H}$

$∴ \: 0.4 = 1 - \frac{300}{T _{ H}} →T _{ H} = 500k$

Now the efficiency of the engine increased by 50% of original efficiency

∴ New efficiency

$η'= 40\% + 20\% = 60\%$

$∴ \: η' = 1 - \frac {T_{ L } }{T _ H}$

$0.6 = 1 - \frac{300}{ {T _ H} }$

$\frac{300}{ {T _ H} } = 0.4$

${T _ H} = 750k$

increase in source the temperature

$750 - 500 = 250k$