Question of thermal physics:-
An Ideal refrigerator works between 27C to 127C temperature. If it expel 120cal heat in 1sec then find avg power in water.
Answers
Answered by
25
Given:-
- T1 = 127°C
- T2 = 27°C
- Expelled heat = Q1 = 120 Cal
━━━━━━━━━━━━━━━━
Need to find:-
- Power = ?
━━━━━━━━━━━━━━━━
Concept:-
- For Refrigerator Heat released = Heat absorbed + work done
- Power = Work done ÷ Time taken
- Carnot engine formula
- When no relation is given, remember to use the formula of Carnot engine for finding necessary values.
━━━━━━━━━━━━━━━━
T1 = 127°C
➟ T1 = 127 + 273 K
➟T1 = 400K
And
T2 = 27°C
➟T2 = 27 + 273
➟T2 = 300K
We know,
C.O.P.=
Substituting their values we get,
➟
➟
➟
━━━━━━━━━━━
We know,
Q1 = Q2 + Work done
➟ 120 Cal = 90 Cal + Work done
➟ Work done = 120 - 90 Cal
➟ Work done = 30 Cal
➟ Work done = 30 × 4.2 Joules
➟
━━━━━━━━━━━━
Power :-
= Work done ÷ Time taken
= 126 Joules ÷ 1 Sec
= 126 Watt
Ataraxia:
Great! <3
Similar questions
English,
2 months ago
Social Sciences,
2 months ago
Physics,
2 months ago
Social Sciences,
5 months ago
Chemistry,
5 months ago
Chemistry,
11 months ago
English,
11 months ago