How much heat is lost when 2012 g of H2O at 400 K is changed into ice at 263 K?

The amount of lost heat is : - 437.7 kJ

Explanation:

This is a calorimetry excersise in three phases.

- When water decreases the T°, always in liquid state

- When water starts to freeze (change state)

- When the ice finally reaches the final T°

Specific heat of water: 4,18 kJ/kg °K

Latent heat of fusion: 334 kJ/kg

Specific heat of ice: 2.05 kJ/kg °K

First step:

Water decreases T° from 400K to 273K

Q = m. C. ΔT

Q = 2012g. 4,18 kJ/kg K. (273K-400K)

Notice we have mass in grams, and we have kg in the units of specific heat. Let's convert g to kg.

2012 g = 2.012 kg

Q = 2.012kg. 4,18 kJ/kg °K. - 127K

Q = - 1068.7 kJ

Second step:

Water liquid is changing state, to solid.

Q = m. Latent heat

Q = 2.012 kg. 334 kJ/kg = 672.008 kJ

Third step:

Ice has to reach to 263 K

Q = 2.012kg. 2.05 kJ/kg °K. (263K - 273K)

Q = 2.012kg. 2.05 kJ/kg °K. - 10K

Q = - 41.2 kJ

Total heat: - 1068.7 kJ + 672.008 kJ + - 41.2 kJ = - 437.7 kJ