Worked example — From ice to steam

Problem

How much heat is needed to take $50$ g of ice at $-10\ ^{\circ}$C to $50$ g of steam at $120\ ^{\circ}$C, at atmospheric pressure? Data: $c_{\text{ghiaccio}}=2090$, $c_{\text{acqua}}=4186$, $c_{\text{vapore}}=2010\ \mathrm{J/(kg\cdot K)}$; $L_f=3{,}34\cdot10^{5}$, $L_v=2{,}26\cdot10^{6}\ \mathrm{J/kg}$.

Solution

Setup. The path has five stages: heating the ice, melting, heating the water, vaporisation, heating the steam. Mass $m=0{,}050$ kg. $Q_1=m\,c_{gh}\,\Delta T=0{,}050\cdot 2090\cdot 10=1045\ \mathrm{J}$ $Q_2=m\,L_f=0{,}050\cdot 3{,}34\cdot10^{5}=1{,}67\cdot10^{4}\ \mathrm{J}$ $Q_3=m\,c_{acq}\,\Delta T=0{,}050\cdot 4186\cdot 100\approx 2{,}09\cdot10^{4}\ \mathrm{J}$ $Q_4=m\,L_v=0{,}050\cdot 2{,}26\cdot10^{6}=1{,}13\cdot10^{5}\ \mathrm{J}$ $Q_5=m\,c_{vap}\,\Delta T=0{,}050\cdot 2010\cdot 20\approx 2010\ \mathrm{J}$ Total: $Q_{tot}=Q_1+Q_2+Q_3+Q_4+Q_5\approx 1{,}54\cdot10^{5}\ \mathrm{J}\approx 154\ \mathrm{kJ}$ $\ev{Q_{tot}\approx 154\ \mathrm{kJ}}$ $Q_4$ alone accounts for 73% of the total: boiling the water is far more “expensive” than heating it.

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Topics: Termologia Concepts: Calore latente e cambiamenti di stato · Calore specifico e capacità termica Methods: Diagramma temperatura-calore