Question

How much heat energy, in kilojoules, is required to convert 73.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ?

Answer 1

Concepts and reason

Enthalpy of fusion describes as energy required or released in physical state change of that substance and it is a thermodynamic property. Enthalpies of fusion of the compound or substance are also known as a latent heat of fusion of that compound or substance. Latent heat of fusion is a heat energy required to change state of a specific quantity of the compound or substance from solid to liquid. And latent heat is a equal to the internal energy of the substance or compound and the product of its pressure and volume.

Fundamentals

For the enthalpy change of the water from solid to liquid state, we need enthalpy of fusion that also called latent heat of fusion and the heat required or released from increasing in temperature of substance.

Where, q = heat energy released or required, m = mass of substance, = Specific heat of that substance and = change in temperature.

Specific heat of water = 4.18 J/(g·K) and specific heat of ice = 2.10 J/(g.K) and the enthalpy of fusion (ice to liquid) = 333.55 J/g (heat of fusion of ice) and the mass of water is 73.0 gram.

There are 3 stages: first heat required to increase temperature of ice -18.0 to 0.0 and second find energy required in state change of ice to liquid and third is energy required to increase temperature of liquid water 0.0 to 25.0

Standard specific values are in Kelvin temperature so first changing Celsius to Kelvin.

0.0 so, 25.0

from

Specific enthalpy of fusion of ice to liquid at 0.0 = 333.55 J/g

And for the 73.0 g ice:

Now, we find energy required for the increase temperature of liquid water from 0.0to 25.0

Now, total energy required in all process:

Ans:

Heat energy required to convert 73.0 g of ice at −18.0 to water at 25.0 = 34.737 kJ