ThermodynamicsHard
Question
A refrigerator with coefficient of performance 1/3 releases 200 J of heat to a hot reservoir, then the work done on the working substance is :-
Options
A.
Joule
Joule B.100 Joule
C.
Joule
Joule D.150 Joule
Solution
β = 
Q2 = 50 Joule
W = Q1 - Q2 = 200 - 50 = 150 Joule
Q2 = 50 Joule
W = Q1 - Q2 = 200 - 50 = 150 Joule
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
Cp - Cv = R correctly applicable to which of the following :-...If one mole of a monoatomic gas (γ = 5 / 3 ) is mixed with one mole of a diatomic gas (γ = 7 / 5 ), the value ...A gas undergoes an adiabatic process in which pressure becomes times and volume become of initial volume. If initial abs...Two monoatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept a...A gas takes part in three thermal processes in which it is heated from the same initial state to the same final temperat...