Refreshing drink

For the participants of a meeting a refreshing drink had to be prepared. The mineral water however had in summer the temperature of the room, i.e. 30 °C, for drinking water with a temperature of 10 °C would be suitable. Therefore the secretary took ice from the refrigerator, which after a while had a temperature of exactly 0 ° C and started to melt. To melt 1 kg of ice, heat of 330 kJ is needed, and thus arises water of the same mass, but also of the same temperature. How much ice must the secretary put into the container, where there are 2 litres of mineral water?

Final Answer:

m₂ =  0.4516 kg

Step-by-step explanation:

$$\begin{gathered} t_1=30\text{°C} \\ {t}_2=10\text{°C} \\ {t}_3=0\text{°C} \\ l=330\cdot 1000=330000\text{J/kg} \\ V=2\text{l}\to {\text{dm}}^3=2\cdot 1{\text{dm}}^3=2{\text{dm}}^3 \\ \rho =1{\text{kg/dm}}^3 \\ c=4200\text{J/kg/K} \\ {m}_1=\rho \cdot V=1\cdot 2=2\text{kg} \\ {Q}_1=Q_2+L \\ {Q}_1=m_1\cdot c\cdot (t_1-t_2)=2\cdot 4200\cdot (30-10)=168000\text{J} \\ {Q}_1=m_2\cdot c\cdot (t_2-t_3)+m_2\cdot l \\ {m}_2=\frac{Q_1}{c\cdot (t_2-t_3)+l}=\frac{168000}{4200\cdot (10-0)+330000}=0.4516\text{kg} \end{gathered}$$

Try another example