A certain refrigerator is rated as being 39% as efficient as a Carnot refrigerator. To remove 102 J of heat from the interior at 0 °C and eject it to the outside at 23 °C, how much work must the refrigerator motor do?

To determine the work required by the refrigerator motor, we need to use the formula for the efficiency of a refrigerator, which is given by:

Efficiency = 1 - (Tc/Th),

where Tc is the cold temperature (in Kelvin) and Th is the hot temperature (in Kelvin).

Now, let's calculate the temperatures in Kelvin:

Tc = 0 °C + 273.15 °K = 273.15 °K,
Th = 23 °C + 273.15 °K = 296.15 °K.

Since the given refrigerator is 39% as efficient as a Carnot refrigerator, we have:

Efficiency = 0.39.
Therefore, 0.39 = 1 - (Tc/Th).

Now, we can solve for Tc/Th:

Tc/Th = 1 - 0.39,
Tc/Th = 0.61.

Next, we can use the relationship between the amount of heat removed (Qc) from the interior and the work done (W) by the motor:

Qc = W * (Th - Tc)/Th.

We are given that Qc = 102 J. Plugging in the known values, we get:

102 J = W * (296.15 °K - 273.15 °K)/296.15 °K.

Now, solve for W:

W = (102 J) * (296.15 °K/23 °K),
W ≈ 1320 J.

Therefore, the refrigerator motor must do approximately 1320 J of work.