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Question: It is given that the temperature on the inside and outside of a refrigerator are \( 275K \) and \( 3...

It is given that the temperature on the inside and outside of a refrigerator are 275K275K and 300K300K . In case we take the cycle of the refrigerator to be ideally reversible, then calculate the heat emitted to the surroundings for every joule of work done?
(a) 10J10J
(b) 11J11J
(c) 14J14J
(d) 12J12J

Explanation

Solution

Hint : We can define an ideally reversible process to be one where there is no change made in the system or its surroundings, that is it is more efficient.
We must remember the relationship between the temperatures of the refrigerator, energy of system and work done;
Q2W=T2T1T2\Rightarrow \dfrac{{Q_2}}{W} = \dfrac{{T_2}}{{T_1 - T_2}}

Complete Step By Step Answer:
Let us initially note down all the given data;
Let us assume that the temperature outside the refrigerator is T1T_1 and the temperature inside the refrigerator is T2T_2 , then the given data says;
\Rightarrow T1=300K T2=275K  T_1 = 300K \\\ T_2 = 275K \\\
Also given that work done is;
W=1J\Rightarrow W = 1J
Now remember the relation between the temperatures of the refrigerator, energy of system and work done;
Q2W=T2T1T2\Rightarrow \dfrac{{Q_2}}{W} = \dfrac{{T_2}}{{T_1 - T_2}}
By substituting all the given values in this equation, we can find the value of Q2Q_2 ;
Q21J=275K(300275)K\Rightarrow \dfrac{{Q_2}}{{1J}} = \dfrac{{275K}}{{(300 - 275)K}}
Q2=275K×1J25K\Rightarrow Q_2 = \dfrac{{275K \times 1J}}{{25K}}
Q2=11J\Rightarrow Q_2 = 11J
Now we have found out the energy of the system to be Q2=11JQ_2 = 11J
We need to find the heat that is emitted to the surrounding which is; Q1Q_1
But for that we need to know the total work done is the difference between heat of surrounding and that of the system;
Q1Q2=W\Rightarrow Q_1 - Q_2 = W
From that equation we can substitute values of WW and Q2Q_2 ;
Q1=(1+11)J\Rightarrow Q_1 = (1 + 11)J
Q1=12J\Rightarrow Q_1 = 12J of heat is delivered to the surrounding
So option (D) 12J12J , is the final answer.

Note :
During the early 19th19th century, it was quite common to use steam engines. The mechanism of steam engines seemed to be as follows, that is by converting heat or thermal energy to motive power, but there was no systematic approach to do this back then. In 18241824 a book written by Nicolas Sadi Carnot, was the first time the idealized model for steam engines, that is of a thermodynamic system was introduced and that is the Carnot cycle process we have today. The cycle comprises four processes; (i) a reversible isothermal gas expansion process, (ii) a reversible adiabatic gas expansion process, (iii) a reversible isothermal gas compression process and (iv) a reversible adiabatic gas compression process.