Question

How much heat is required to convert solid ice with a mass of 605 g and at a temperature of -12.5 °C to liquid water at a temperature of 55.5 °C? (The specific heat of ice is c_ice = 2100 J/kgK, the specific heat of water is c_water = 4186.8 J/kgK, and the heat of fusion for water is: L_f = 334 kJ/kg.)

Answer 1

we have

mass of the ice, m = 605g = 0.605 kg

specific heat capacity of ice, c_ice = 2100 J / kg K

specific heat capacity of water, c_water = 4186.8 J / kg K

latent heat of fusion of ice, L_{f (ice)} = 334 kJ / kg

We can divide the entire process into three different process (just for calculation)

Q₁ = heat required to convert ice at –12.5 °C to ice at 0 °C = m c_ice ΔT₁

= 0.605 kg * 2100 J / kg K * {0 - (-12.5)} °C

= 15881.25 J

= 15.88 kJ

We are taking the difference between two temperatures in ΔT, therefore, the unit conversion from Celsius to Kelvin will not have any impact on magnitude. Obviously, the units will cancel out.

Q₂ = heat required to melt ice at 0 °C to water at 0 °C = mL_{f (ice)}

= 0.605 kg * 334 kJ / kg

= 202.07 kJ

Q₃ = heat required to convert water at 0 °C to water at 55.5 °C = m cwater ΔT₂

= 0.605 kg * 4186.8 J / kg K * (55.5 - 0) °C

= 140582.277 J

= 140.58 kJ

Q = Total heat required to convert ice at -12.5 °C to water at 55.5 °C = Q₁ + Q₂ + Q₃

= (15.88 + 202.07 + 140.58) kJ

= 358.53 kJ