1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A Carnot heat engine works between the temperatures 427∘C and 27∘C. What amount of heat should it consume per second to deliver mechanical work at the rate of 1.0 kW?

A
418 cal/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
41.8 cal/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4.18 cal/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.418 cal/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 418 cal/sThe efficiency of the carnot heat engine is η=1−T2T1 ...(i) T2=(273+27)=300 K T1=(273+427)=700 K ⇒η=1−(300700)=47 Also the efficiency is defined as, η=WQ1 ...(ii) ⇒Q1=Wη=1.0 kW4/7=1.75 kJ/s since 1 cal=4.186 J ∴Q1=1.754.186=0.418 kcal/s=418 cal/s Thus, the engine would require 418 cal of heat per second, to deliver the requisite amount of work.

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program