Question

If efficiency of a carnot engine is 20 and difference between its two source is 80 K then temperature of its heat source is____

* 410K
* 400K
* 300K
* 273K​

Answer 1

400k is the correct answer

Answer 2

Let temperature of sink be T , so temperature of source will be T + 80 .

Now, we know that :

$\eta = 1 - \dfrac{T_{sink}}{T_{source}}$

$\implies \eta = 1 - \dfrac{T}{T + 80}$

$\implies 20\% = 1 - \dfrac{T}{T + 80}$

$\implies 0.2 = 1 - \dfrac{T}{T + 80}$

$\implies \dfrac{T}{T + 80} = 0.8$

$\implies 100T = 80T + 6400$

$\implies 20T = 6400$

$\implies T = 320$

So, temperature of source will be :

$T_{source} = 320 + 80 = 400K$

Option B) is correct !