Question

A carnot engine has an efficiency 25%. It operates b/w reservoirs of constant temperatures with temp. diff. of 80 degree celsius. what is the temp.of the low temp.reservoir...

its urgent plsss help

Answer 1

80 because the temperature is constant

Answer 2

The temperature is 240 degree Celsius.

Given:

Efficiency is Carnot engine = 25%

Temperature difference = 80 degree Celsius = 80

Solution:

The efficiency of a Carnot engine is given by the formula given below:

$Efficiency = 1-\left(\frac{t_{2}}{t_{1}}\right)$

$25 \% = 1-\left(\frac{t_{2}}{t_{1}}\right)$

$\frac{25}{100} = \frac{t_{1}-t_{2}}{t_{1}}$

$\frac{25}{100} = \frac{80}{t_{1}}$

$t_{1} = \frac{80 \times 100}{25}$

$t_{1} = \frac{8000}{25}$

$t_{1} = 320 \ degree \ Celsius$

So, the temperature reservoir 1 has the temperature of 320 degree Celsius

Thus, the temperature of other reservoir is given below:

$\eta=1-\frac{t_{2}}{t_{1}}$

$0.25=\frac{t_{1}-t_{2}}{t_{1}}$

$0.25=\frac{320-t_{2}}{320}$

$320-t_{2}=(320 \times 0.25)$

$320-t_{2}=80$

$t_{2}=240 \ degree \ Celsius$