How much heat is required to convert 18.0 g of ice at - 10.0C to steam at 100.0C? Express your answer in joules, calories, and Btu.

Amount of heat required = 54601.2 J

Amount of heat required = 13050 cal

Amount of heat required = 51.68 Btu

Explanation:

Mass of ice, m = 18.0 g

Initial temperature of ice, T₀ = - 10.0 ⁰C

Specific heat of ice, C₀ = 0.50 cal/g-°C

Final temperature of ice, T₁ = 0 ⁰C

Amount of heat required to change the temperature of ice from T₀ to T₁ is:

Q₁ = mC₀ (T₁ - T₀)

Q₁ = 18 x 0.50 x (0 + 10)

Q₁ = 90 cal

Latent heat of ice, L₁ = 80 cal/g

Amount of heat required to change ice into water at T₁ temperature is:

Q₂ = m x L₁

Q₂ = 18 x 80 = 1440 cal

Final temperature of water, T₂ = 100 °C

Specific heat of water, C₁ = 1 cal/g-°C

Amount of heat required to change the temperature of water from 0 °C to 100 °C, that is, from T₁ to T₂ is:

Q₃ = mC₁ (T₂ - T₁)

Q₃ = 18 x 1 x (100 - 0)

Q₃ = 1800 cal

Latent heat for boiling, L₂ = 540 cal/g

Amount of heat required to change water into steam at 100 °C is:

Q₄ = mL₂

Q₄ = 18 x 540 = 9720 cal

Total amount of heat required to change ice at - 10 °C to steam at 100 °C is:

Q = Q₁ + Q₂ + Q₃ + Q₄

Q = 90 + 1440 + 1800 + 9720

Q = 13050 cal

But, 1 cal = 4.184 joule

So, in joules the heat required is:

Q = 13050 x 4.184 = 54601.2 J

1 cal = 3.96 x 10⁻³ Btu

In terms of Btu, the heat required is:

Q = 13050 x 3.96 x 10⁻³ = 51.68 Btu