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: Determine the mass of melted ice

The total mass of ice is \(m = 450\space g\). The fraction of ice that melts is \(\frac{2}{3}\). So the mass of melted ice \(m_{melt}=\frac{2}{3}\times450\space g = 300\space g=0.3\space kg\)

Step2: Use the formula for heat of fusion

The heat of fusion for water \(L_f = 3.34\times 10^{5}\space J/kg\). The formula for the heat absorbed \(Q = m_{melt}L_f\)
Substitute \(m_{melt}=0.3\space kg\) and \(L_f = 3.34\times 10^{5}\space J/kg\) into the formula:
\(Q=(0.3\space kg)\times(3.34\times 10^{5}\space J/kg)\)
\(Q = 1.002\times 10^{5}\space J\)

Answer:

\(1.002\times 10^{5}\space J\)