Question

a carnot engine having efficiency 1/5 has source temperature pf 527°c . what must be change in sink temperature so that efficiency becomes 1/8​

Answer 1

Answer:

The change in sink temperature is 60 K

Explanation:

Given :

A Carnot engine having efficiency 1/5 has source temperature of 527°C

To find :

the change in sink temperature so that efficiency becomes 1/8

Solution :

The formula for efficiency of Carnot Engine is given by,

   $\underline{\boxed{\bf \eta =1-\dfrac{T_2}{T_1}}}$

where

T₂ denotes the temperature of sink

T₁ denotes the temperature of source

Case - (i) :

efficiency, η = 1/5

T₁ = 527°C = 527 + 273 K = 800 K

 $\sf \dfrac{1}{5} = 1-\dfrac{T_2}{800} \\\\ \sf \dfrac{T_2}{800} =1-\dfrac{1}{5} \\\\ \sf \dfrac{T_2}{800} =\dfrac{5-1}{5} \\\\ \sf \dfrac{T_2}{800} = \dfrac{4}{5} \\\\ \sf T_2= \dfrac{4}{5} \times 800 \\\\ \sf T_2=640 K$

The temperature of the sink when efficiency is 1/5 is 640 K

Case - (ii) :

efficiency, η = 1/8

T₁ = 800 K

$\sf \dfrac{1}{8} = 1-\dfrac{T_2}{800} \\\\ \sf \dfrac{T_2}{800} =1-\dfrac{1}{8} \\\\ \sf \dfrac{T_2}{800} =\dfrac{8-1}{8} \\\\ \sf \dfrac{T_2}{800} = \dfrac{7}{8} \\\\ \sf T_2= \dfrac{7}{8} \times 800 \\\\ \sf T_2=700 K$

The temperature of the sink when efficiency becomes 1/8 is 700 K

The change in sink temperature :

= 700 K - 640 K

= 60 K