An insulated beaker with negligible mass contains liquid wat

An insulated beaker with negligible mass contains liquid water with a mass of 0.340 kg and a temperature of 73.5 C .

1. How much ice at a temperature of -15.1 C must be dropped into the water so that the final temperature of the system will be 40.0 C ?

Take the specific heat of liquid water to be 4190 J/kgK , the specific heat of ice to be 2100 J/kgK , and the heat of fusion for water to be 3.34×10⁵ J/kg .

Solution

0.34 kg * 4190 J / kg K * (73.5 - 40) = 47724.1 J

needs to be absorbed by the ice

m kg of ice from -15.1 C to 40 C water will absorb =

m * ( 15.1 * 2100 J + 334000 J + 40 * 4190 J ) = m*533310 J

47724.1 J = m*533310 J

m = 47724.1/533310 = 0.0895 kg

mass of ice = 89.5 gram