Question

a carnot engine whose sink is at 300 K has an efficiency of 50 percent.by how much should the temperature of source be increased so as the efficiency becomes 70 percent​

Answer 1

$\boxed{\sf \green{Answer}}$

400k or 400 °C

$\boxed{\sf \green{Explanation}}$

Given

• efficiency % $(\eta = 50 )$
• sink temp. (T2) = 300k

To Find:

Increase in the Temp of the source, when efficuency becomes 70%

Solution:

Temp. of source when efficiency is 50%

w.k.t

$\boxed{\bold {\pink{\eta = (1 - \dfrac{T_2}{T1}) \times 100}}}$

$\mathsf{50 = (1 - \dfrac{300}{T_1x}) \times 100}$

$\mathsf{1 = (1 - \dfrac{300}{T_1}) \times 2}$

$\mathsf{1= ( \dfrac{T_1 - 300}{T_1}) \times 2}$

$\mathsf{T_1= 2T_1 - 600}$

$\mathsf{T_1= 600}$

Temp. of source when efficiency is 70%

$\mathsf{70 = (1 - \dfrac{300}{T_1'}) \times 100}$

$\mathsf{7= (\dfrac{T_1'- 300}{T_1'}) \times 10}$

$\mathsf{7T_1'= 10T_1'- 3000}$

$\mathsf{3T_1'= 3000}$

$\mathsf{T_1' = 1000}$

Increase in the temp. of source

$\mathsf{T_1' -T_1 = 1000 - 600}$

$\mathsf{T_1' -T_1 = 400k\: or \: 400\degree C}$