In a vapour absorption refrigerator, the temperatures of evaporator and ambient air are 10°C and 30°C respectively. For obtaining COP of 2 for this system, the temperature of the generator is to be nearly

This question was previously asked in
ESE Mechanical 2018 Official Paper
View all UPSC IES Papers >
  1. 90 °C
  2. 85 °C
  3. 80 °C
  4. 75 °C

Answer (Detailed Solution Below)

Option 3 : 80 °C
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
20 Qs. 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

Concept:

TG is the generator temperature, TE is the evaporator temperature and T0 is the environment temperature

Calculation:

Given: 

TE = 10°C = 283 K, T0 = 30°C = 303 K, COP = 2

Now,

TG = 352.87 K

TG = 79.87°C 

Latest UPSC IES Updates

Last updated on Jun 23, 2025

-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.

->  UPSC ESE admit card 2025 for the prelims exam has been released. 

-> The UPSC IES Prelims 2025 will be held on 8th June 2025.

-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.

-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.

More Vapour Absorption (V-A) Cycle Questions

Hot Links: teen patti vungo teen patti download apk teen patti gold new version 2024 mpl teen patti teen patti gold old version