Latent heat

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

Final Answer:

Q =  45.56 kJ

Step-by-step explanation:

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

Try another example