91Ó°ÊÓ

Mass of water: 150 kg

Temperature of water: $28°C$

Mass of ice: 0.600 kg

Temperature of ice: $-22°C$

When two systems are connected to each other via a path that permits heat flow. A point at which the heat flow stops is called as thermal equilibrium

Heat from water when cooled at $0°C$is given by,

$Q=mc∆T$

Here, m is the mass of water, c is the specific heat and $∆T$is the change in temperature.

Substitute values in the above,

$$\begin{gathered} Q=\left(1.5kg\right)\left(4190J/kg.K\right)\left(28K\right) \\ =1.76×{10}^{5}J \end{gathered}$$

Ice will melt by the hear given,

$Q=mc∆T+m{L}_f$

Here, m is the mass of water, c is the specific heat, is the change in temperature and is the latent heat of ice.

Substitute all the values in the above,

$$\begin{gathered} Q=\left(0.600kg\right)\left[\left(2100J/kg.K\right)\left(22K\right)+3.34×{10}^{5}J/kg\right] \\ =2.276×{10}^{5}J \end{gathered}$$

Thus, the heat required is $2.276×{10}^{5}J$.

For the melting of ice,

$$\begin{gathered} Q_{fromwater}=mc∆T+m_{ice}{L}_f \\ {m}_{ice}=\frac{Q_{fromwater}-mc∆T}{L_f} \end{gathered}$$

Here, m is the mass of water, c is the specific heat, $∆T$is the change in temperature and $L_f$is the latent heat of ice.

Substitute all the values in the above,

$$\begin{gathered} m_{ice}=\frac{1.76×{10}^{5}J-\left(0.600kg\right)\left(2100J/kg.K\right)\left(22K\right)}{3.34×{10}^{5}J/kg} \\ {m}_{ice}=0.444kg \end{gathered}$$

The ice remaining is given by,

$$\begin{gathered} m_{remaining}=0.600kg=0.444kg \\ =0.156kg \end{gathered}$$

Thus, there will be still ice remaining in the water so the final temperature is $0°C$.