Answer
$m_{ice}=0.261\ lbm$
Work Step by Step
Energy balance for the can:
$Q=\Delta U=mc\Delta T$
Properties of water at an average temperature:
$\rho=62.3\ lbm/ft³,\ c_p=1.00\ Btu/lbm.°R$
Since $V_{can}=12\ oz=0.0125\ ft³$
Therefore $m_{can}=0.781\ lbm$
Hence $Q_{can}=-37.5\ Btu$
With $\Delta h_{fusion}=143.5\ Btu/lbm$:
$Q_{ice}=-Q_{can}=m_{ice}\Delta h_{fusion}$
Solving for the mass:
$m_{ice}=0.261\ lbm$
