Question

How much heat energy, in kilojoules, is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ?

Answer 1

Given:
Specific heat of ice: \( C_{s, ice}=2.09\ J \ g^{−1} °C^{−1}\)
Specific heat of liquid water: \(C_{s, water}=4.18 J g^{−1} ∘C^{−1}\)
Enthalpy of fusion:\( Δ_{fus}H∘=334 J g^{−1}\)
Enthalpy of vaporization:\( Δ_{vap}H∘=2250 J g^{−1}\)

Answer 2

You need to add three values together to find the answer:

1. The amount of heat required to raise the temperature of ice from -18.0 C to 0 C, which is the freezing/melting point.
2. The amount of heat required to convert the ice from a solid into a liquid (i.e. water).
3. The amount of heat required to raise the temperature of the water from 0 C to 25 C.

Calculate each value separately and find the sum for your answer!

Answer 3

Okay, this is how I did it:

q1 will be the heat to raise the temp of ice from -18 to zero
q1= 53(2.09)(0-(-18)) = 1993.86

then q2 is the heat to convert ice to liquid water
q2 = 53 (334) = 17702

and the last one, q3 which is the amt of heat for water from 0 to 25 celsius:
q3=53(4.18)(25-0)=5538.5

adding them together, i'll get 25234.36 J
ooh I got it, thank you so much!