Question

1. 65.0 kilojoules of energy are added to 150 g of ice at 0.0°c. what is the final temperature of the water?

Explanation:

Step1: Calculate heat for phase change

The heat of fusion of ice \(H_f = 334\space J/g\).
For \(m = 150\space g\) of ice, the heat required to melt it \(Q_{melt}=m\times H_f\)
\(Q_{melt}=150\times334 = 50100\space J\)

Step2: Calculate remaining heat

Total heat \(Q_{total}=65000\space J\)
Remaining heat \(Q_{remain}=Q_{total}-Q_{melt}\)
\(Q_{remain}=65000 - 50100=14900\space J\)

Step3: Calculate temperature change of water

The specific heat capacity of water \(c = 4.18\space J/(g\cdot^{\circ}C)\)
Using \(Q = mc\Delta T\), where \(\Delta T=T - 0\) (initial temperature of melted ice is \(0^{\circ}C\))
\(\Delta T=\frac{Q_{remain}}{mc}\)
Substitute \(m = 150\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\) and \(Q_{remain}=14900\space J\)
\(\Delta T=\frac{14900}{150\times4.18}\approx23.7^{\circ}C\)

Answer:

The final temperature of the water is approximately \(23.7^{\circ}C\)