Question

1. suppose a bag full of ice (450 g) at 0.0 °c sits on the counter and begins to melt to liquid water. how much energy must be absorbed by the ice if 2/3 of it melted?

Explanation:

Step1: Calculate the mass of melted ice

The mass of the ice is \(m = 450\space g\). The fraction of ice that melted is \(\frac{2}{3}\). So the mass of melted ice \(m_{melted}=450\times\frac{2}{3}\space g = 300\space g\)

Step2: Use the formula for heat of fusion

The heat of fusion of ice \(H_f=334\space J/g\). The formula for the energy absorbed \(Q = m\times H_f\). Substitute \(m = 300\space g\) and \(H_f = 334\space J/g\) into the formula: \(Q=300\times334\space J\)

Step3: Calculate the value of \(Q\)

\(Q = 300\times334=100200\space J = 100.2\space kJ\)

Answer:

\(100.2\space kJ\)