How much heat (in kj) is required to warm 13.0 g of ice, initially at -12.0 ∘c, to steam at 113.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘
c.

39.82693 kJ

Explanation:

Sensible heat is computed as follows:

Q = m*cp*ΔT

where Q is heat, m is mass, cp is heat capacity and ΔT is temperature change.

Latent heat transformation is computed as follows:

Q = m*λ

where λ is latent heat

water latent heat of fusion: 334 J/g

water latent heat of vaporization: 2260 J/g

First you have to heat ice from -12 °C to its melting point (0 °C) as follows:(this is a sensible heat transformation)

Q1 = m*cp*ΔT

Q1 = 13*2.09*(0 - (-12))

Q1 = 326.04 J = 0.32604 kJ

Then you have to melt the ice into water as follows:

(this is a latent heat transformation)

Q2 = m*λ

Q2 = 13*334

Q2 = 4342 J = 4.342 kJ

Next, you have to heat the water from 0 °C to its boiling point (100 °C) as follows:

(this is a sensible heat transformation)

Q3 = m*cp*ΔT

Q3 = 13*4.184*(100 - 0)   (heat capacity of water: 4.184 J/(g °C))

Q3 = 5439.2 J = 5.4392 kJ

Then you have to evaporate the water into steam as follows:

(this is a latent heat transformation)

Q4 = m*λ

Q4 = 13*2260

Q4 = 29380 J = 29.38 kJ

Finally, you have to heat the steam from 100 °C to 113 °C as follows:

(this is a sensible heat transformation)

Q5 = m*cp*ΔT

Q5 = 13*2.01*(113 - 100)

Q5 = 339.69 J = 0.33969 kJ

Total heat, Q = Q1 + Q2 + Q3 + Q4 + Q5 = 0.32604 + 4.342 + 5.4392 + 29.38 + 0.33969 = 39.82693 kJ

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