This can be solve using heat
balance. The amount of heat the lead released is equal to the amount of heat
the ice absorb.
The heat release by the leas = m1CpdeltaT
Where m1 is the mass of lead
Cp is the specific heat of lead
Delta is the change of temperature
Heat absorbed by the ice = m2Lf
Where m2 is the mass of ice
Lf is the latent heat of fusion of
ice
M1CpdeltaT = m2Lf
(15g) ( 0.128 j/g C) ( 120 -0 C) =
m2(333/g)
Solving for m2
M2 = 0.69 kg of ice melts
