# 900 kg water have T1 = -40 C, T2 = 0C and T3 = 5 C, How much energy is needed to melt this mass of water in 2 hours? Thank you so much.

Jul 16, 2016

You will need to use 395 MJ of energy. The amount of time is immaterial.

#### Explanation:

There are three heat transfers involved.

$q = \text{heat to warm ice" + "heat to melt ice" + "heat to warm water}$

$q = {q}_{1} + {q}_{2} + {q}_{3}$

The specific heat capacity of ice is $\text{2090 J·kg"^"-1""°C"^"-1}$

q_1 = mcΔT = 900 color(red)(cancel(color(black)("kg"))) ×" 2090 J"·color(red)(cancel(color(black)("kg"^"-1""°C"^"-1"))) × 40 color(red)(cancel(color(black)("°C"))) = "75 200 000 J" = "75.2 MJ"

The heat of fusion of ice is 334 kJ/kg.

q_2 =mΔ_"fus"H^° = 900 color(red)(cancel(color(black)("kg"))) × "334 kJ"·color(red)(cancel(color(black)("kg"^"-1"))) = "300 600 kJ" = "300.6 MJ"

The specific heat capacity of water is $\text{4181 J·kg"^"-1""°C"^"-1}$.

q_3 = mcΔT = 900 color(red)(cancel(color(black)("kg"))) × "4181 J"·color(red)(cancel(color(black)("kg"^"-1""°C"^"-1"))) × 5 color(red)(cancel(color(black)("°C"))) = "19 000 000 J" = "19 MJ".

$q = {q}_{1} + {q}_{2} + {q}_{3} = \text{(75.2 + 300.6 + 19) MJ" = "395 MJ}$