Question

A refrigerator removes heat from the freezing compartment at the rate of20 kJ per cycle and ejects 24 kJ into the room each cycle. How much energy is used in each cycle?

Answer 1

The refrigerator uses 4 kJ of energy each cycle, calculated by the work done W, which is the difference between the heat ejected Qh (24 kJ) and the heat removed Qc (20 kJ).

Step-by-step explanation:

The question is calculating the energy used by a refrigerator in each cycle. The refrigerator absorbs heat Qc from the inside and ejects a larger amount of heat Qh to the room. The extra energy being ejected comes from the work W done by the refrigerator, which is the energy used by it in each cycle.

If the refrigerator removes 20 kJ (Qc) from the freezing compartment and ejects 24 kJ (Qh) into the room each cycle, the work done (energy used) W can be found using the first law of thermodynamics which states that the conservation of energy principle - the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

W = Qh - Qc

Substituting in the given values:

W = 24 kJ - 20 kJ

W = 4 kJ

Thus, the work done (energy used) by the refrigerator in each cycle is 4 kJ.

Answer 2

Energy used = 4KJ

Step-by-step explanation:

Second law of thermodynamics states that as energy is transferred or transformed, more and more of it is wasted. The Second Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state.

Now when we apply that to heat engines, we'll see that;

Heat expelled = Heat removed + Work Done

We can write it as;

Q_h = Q_c + W

We are given that;

Heat removed; Q_c = 20KJ

Heat expelled into the room in each cycle; Q_h = 24KJ

Thus; plugging these 2 values into the equation, we obtain;

24 = 20 + W

W = 24 - 20

W = 4 KJ

Work done is energy used.

Thus, energy used = 4 KJ