An insulated beaker with negligible mass contains liquid wat

An insulated beaker with negligible mass contains liquid water with a mass of 0.315 kg and a temperature of 82.4 degree C. How much ice at a temperature of -18.1 degree C must be dropped into the water so that the final temperature of the system will be 28.0 degree C? Take the specific heat of liquid water to be 4190 J\\kg. K, the specific heat of ice to be 2100 J\\kg. K, and the heat of the fusion for water to be 3.34x10^5 J\\kg.

Solution

Heat lost by water is equal to heat gained by ice. that is,

MC_wT_w= m [C_iceT_ice+ L +C_wT )

Amount of ice that must be added is

m = MC_wT_w/[ C_iceT_ice+ L +C_wT ]

Cw = 4190,

Tw = (82.4 -28) =54.4

Cice = 2100,

Tice = 18.1,

L = 334000 ,

M =0.315,

T = 28

therefore, the mass m is, m = 0.14673 kg = 0.147 kg