How much energy (heat) is required to convert 52.0 g of ice at - 10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Specific heat of water 4.18 J/g • °C Specific heat of steam 1.84 J/g • °C Molar heat of fusion 6.02 kJ/mol Molar heat of vaporization 40.7 kJ/mol

Answer: The energy (heat) required to convert 52.0 g of ice at - 10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from - 10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

q = 17.38 kJ

And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

Add up all the energy gives 157.8 kJ