Suppose 400. g of water at 30.0 °C is poured over a 60.0-g cube of ice with a temperature of -5.00 °C. If all the ice melts, what is the final temperature of the water? If all of the ice does not melt, how much ice remains when the water-ice mixture reaches equilibrium?
Express the relation for heat evolved for steam.
\(Q=m_{s} L+m_{s} C_{W}\left(T_{s}-T\right) \ldots \ldots\) (1)
Here, \(\mathrm{Q}\) is heat evolved, \(m_{s}\) is the mass of the steam, \(\mathrm{L}\) is the latent heat of the steam, \(C_{W}\) is the specific capacity of water, \(T_{s}\) is the temperature of the steam and \(T\) is the final temperature.
Express the relation for heat evolved for water and aluminium.
$$ Q=m_{W} C_{W}\left(T-T_{W}\right)+m_{A l} C_{A l}\left(T-T_{A l}\right) \cdots \cdots $$
Here, \(m_{A l}\) is the mass of the aluminium, \(C_{A l}\) is the specific heat capacity of aluminium, \(T_{A l}\) is
the temperature of aluminium, \(m_{W}\) is the mass of the water, and \(T_{W}\) is the temperature of water.
Use equation (1) and equation (2) to compare the expressions.
$$ m_{s} L+m_{s} C_{s}\left(T_{s}-T\right)=m_{W} C_{W}\left(T-T_{W}\right)+m_{A} C_{A}\left(T-T_{A}\right) $$
Substitute \(0.010 \mathrm{~kg}\) for \(m_{s}, 2260 \times 10^{3} \mathrm{~J} / \mathrm{kg}\) for \(L, 100^{\circ} \mathrm{C}\) for \(T_{s}, 0.10 \mathrm{~kg}\) for \(m_{W},\)
\(4.18 \times 10^{3} \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\) for \(C_{W}, 35 \mathrm{~kg}\) for \(m_{A l}, 0.900 \times 10^{3} \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\) for \(C_{A l},\) and \(19^{\circ} \mathrm{C}\) for \(T_{W}\) and
\(T_{A l}\)
$$ \begin{aligned} \left[\begin{array}{l} (0.010 \mathrm{~kg}) \times\left(2260 \times 10^{3} \mathrm{~J} / \mathrm{kg}\right) \\ +(0.010 \mathrm{~kg})\left(4.18 \times 10^{3} \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\right)\left(100^{\circ} \mathrm{C}-T\right) \end{array}\right] &=\left[\begin{array}{l} (0.10 \mathrm{~kg})\left(4.18 \times 10^{3} \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\right)\left(T-19^{\circ} \mathrm{C}\right) \\ 2,260+41,800-418 T & =41.8 T-7,942+31,500-598,500 \\ 31583 T & =650502 \\ & =2.05^{\circ} \mathrm{C} \end{array}\right. \end{aligned} $$
Therefore, final temperature is \(2.05^{\circ} \mathrm{C}\).