How much thermal energy (in kcal) is required to change a 43 g ice cube from a solid at - 16.5 oC to steam at 11.5 oC above boiling

The total thermal energy required  is 8.93 Kcal

Explanation:

Given;

mass of the ice cube, m = 43 g

specific heat capacity of water, Cp = 4.18 J/gc

specific latent heat of fusion of ice, Cf = 334 J/g

First step, determine the heat needed to raise the temperature of the ice from -16.5 °C to 0° C

Q₁ = mCp[0 - (-16.5)]

Q₁ = 43 x 4.18(16.5)

Q₁ = 2965.71 J

Second step, determine the latent heat of fusion of ice at 0°C

Q₂ = mCf

Q₂ = 43 x 334

Q₂ = 14362 J

Third step, determine the quantity of heat required to raise the temperature of the water initially at 0°C to above 11.5 °C of boiling point of water.

The final temperature of the water = 11.5 °C + 100° C = 111.5 °C

Q₃ = mCp Δθ

Q₃ = 43 x 4.18 (111.5 - 0)

Q₃ = 20041 J

Total thermal energy required = Q₁ + Q₂ + Q₃

Total thermal energy required =  2965.71 J + 14362 J +  20041 J

Total thermal energy required = 37,368.71 J

Total thermal energy required = 8.93 Kcal

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