The heat of vaporization of a certain refrigerant is 148 J/g. Calculate the number of kilograms of refrigerant that must be evaporated to freeze a tray of 16 one-ounce (1 oz = 28 g) ice cubes starting with the water at 15°C. Heat capacity (water) = 4.184 J/g °C
\(\Delta\H_fusion (water) = 333 J/g

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat required to freeze the water in the ice cubes:

Q = (16 x 28 g) x [(0°C - 15°C) x 4.184 J/g°C + 333 J/g] = 7,339 J

This is the amount of heat that needs to be absorbed by the refrigerant to freeze the ice cubes. We can use the heat of vaporization to calculate the amount of refrigerant needed:

148 J/g x m = 7,339 J

m = 7,339 J / 148 J/g = 49.5 g

Finally, we convert to kilograms:

m = 49.5 g / 1000 = 0.0495 kg ≈ 0.05 kg

So, the answer is B, 1.2 kg.

The Heat of Vaporization of a Certain Refrigerant Is 148 \(\Delta\Hfusion (Water) = 333 J/g A) 0

The Heat of Vaporization of a Certain Refrigerant Is 148 \(\Delta\H_fusion (Water) = 333 J/g
A) 0