Question

An ice sculpture is made of 55 kg of ice and is stored at –22˚C. Heat is turned on instead of the cold and evaporates all the ice into steam at a temperature of 115˚C. How much heat was added to the sculpture to create the sauna? (grade 11 physics)

Answer 1

The heat will be necessary to:

1 – increase ice temperature from -22 to 0°C: $Q_1=c_1 mΔt_1$

2 – melt the ice: $Q_2=mλ$

3 – heat up the obtained water from 0 to 100°C: $Q_3=c_2 mΔt_2$

4 – evaporate water: $Q_4=mr$

5 – increase steam temperature from 100 to 115°C: $Q_5=c_3 mΔt_3$

The total heat is the sum of all Qs:

$$Q=Q_1+Q_2+Q_2+Q_3+Q_4+Q_5,\\ Q=m(c_1 Δt_1+λ+c_2 Δt_2+r+c_3 Δt_3 ).$$

In this equation:

$c_1=4200 \text{ J/kg/K}\\ c_2=2108\text{ J/kg/K}\\ c_3=1900\text{ J/kg/K}\\ \lambda=330,000\text{ J/kg}\\ r=2,300,000\text{ J/kg}.$

With these numbers the total heat will be about 172 MJ.