The temperature of the reservoir of Carnot’s engine operating with an efficiency of $70$ % is $1000 K$ . The temperature of the source is?

300 k

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

200 k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500 k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $300 k$
For a carnot engine
$η = 1 - \frac{T_{L}}{T_{H}} \frac{70}{100} = 1 - \frac{T_{L}}{1000} \Rightarrow \frac{T_{L}}{1000} = 1 - \frac{70}{100} \Rightarrow \frac{T_{L}}{1000} = \frac{30}{100} \Rightarrow T_{L} = 300 K$

Suggest Corrections

Similar questions

The temperature of reservoir of Carnot's engine operating with an efficiency of 70% is 1000K. The temperature of its sink is

T_{1} = 500 K

T_{2} = 300 K

100^{o} C

- 23^{o} C

100^{o} C

- 23^{o}

50 %

500 K

60 %

Join BYJU'S Learning Program