Question

1.50 kg of ice (-15ºC) is brought to a water (+37ºC). Specific thermal capacity of ice is 2.09 J/K g and water 4.18 J/K g. Molecular ice melting heat is 6.02 kJ/mol. How to calculate the total amount of heat consumed for the specified change?

Answer 1

From ice (-15 degrees) to ice (0 degrees) than from ice (0 degrees) to water (0 degrees) finally water (at 0 degrees) to water (at 37 degrees). Note that I used kiloJoules per degrees C per kg.

Explanation:

Ice density is 917 kg per cubic meter.
Water density is 1000 kg per cubic meter (I assume it is constant).

If you had 1.5 kg of ice, it means it occupies 1.636 Liters (0.001635 cubic meters) (using density of ice).

From -15 degrees to 0 degrees (for 1 kg of ice) heat necessary is `=2.1 (kJ)/(t degrees C)`

`=1.5times2.1times15 = 47.25` kJ energy necessary.

Melting ice latent heat of fusion is 333 kJ (for 1 kg). It is 499.5 kJ (for 1.5 kg).

Water (4.18 kJ/degrees C) temperature increase for 1 kg of water. But we have 1.636 kg of water.

Energy necessary to heat water (from 0 degrees to 37 degrees) is
`=1.636times37times4.18 = 253` kJ.

Add them up `=47.25+499.5+253 = 799.25` kJ. It is the energy necessary to melt down 1.5 kg ice at -15 degrees and heat up water to 37 degrees.