How much heat must be removed to freeze a tray of ice cubes at 0 C if the water has a mass of 225 g?

Question

Mateo Aguilar · Accepted Answer

Well, if the tray itself has a negligible heat capacity...

...then we ignore it and focus on the ice, which has an unstated enthalpy of fusion of #"6.02 kJ/mol"#. (You should be supplying the data here...)

Hence, the amount of heat that is released at constant pressure is:

#color(blue)(|q_"removed"|) = "6.02 kJ"/cancel"mol" xx cancel"1 mol"/(18.015 cancel"g water") xx 225 cancel"g" = color(blue)"75.2 kJ"#

At constant pressure, #q = DeltaH#. What is the enthalpy change for this process in #"kJ"#?

HINT: read the question wording again. Which direction is the heat flowing, and is the ice the system or surroundings?

Mateo Aguilar · Answer

undefined The amount of heat $ q $ required to freeze a substance can be calculated using the equation:

$$ q = m \cdot \Delta H_f $$

Where:
- $ q $ is the heat absorbed or released (in joules)
- $ m $ is the mass of the substance (in grams)
- $ \Delta H_f $ is the heat of fusion (in joules per gram)

For water, the heat of fusion is $ 334 \, \text{J/g} $.

Given that the mass of water is $ 225 \, \text{g} $, and assuming the temperature change is from $ 0^\circ \text{C} $ to $ 0^\circ \text{C} $ during freezing, the heat required is:

$$ q = 225 \, \text{g} \times 334 \, \text{J/g} = 75150 \, \text{J} $$

So, $ 75150 \, \text{J} $ of heat must be removed to freeze the tray of ice cubes.