Answer:
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
