Latent heat

How much heat is needed to take from 100 g of water at 20°C to cool to ice at -18°C? Mass heat capacity c (ice) = 21kJ/kg / °C; c (water) = 4.19 KJ/kg / °C and the mass group heat of solidification of water is l = 334kJ/kg

Correct answer:

Q =  45.56 kJ

Step-by-step explanation:

$$\begin{gathered} m=100\text{ g}\to \text{ kg}=100\mathrm{/}1000\text{ kg}=0.1\text{ kg } \\ {t}_1=20\text{ °C } \\ {t}_2=0\text{ °C } \\ {t}_3=-18\text{ °C } \\ {c}_1=2.1\text{ kJ/kg/°C } \\ {c}_2=4.19\text{ kJ/kg/°C } \\ l=334\text{ kJ/kg } \\ {Q}_1=m\cdot c_2\cdot (t_1-t_2)=0.1\cdot 4.19\cdot (20-0)={\displaystyle \frac{419}{50}}=8.38\text{ kJ } \\ {Q}_2=m\cdot l=0.1\cdot 334={\displaystyle \frac{167}{5}}=33.4\text{ kJ } \\ {Q}_3=m\cdot c_1\cdot (t_2-t_3)=0.1\cdot 2.1\cdot (0-(-18))={\displaystyle \frac{189}{50}}=3.78\text{ kJ } \\ Q=Q_1+Q_2+Q_3=8.38+33.4+3.78=45.56\text{ kJ} \end{gathered}$$

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