Question

Adding Ice to Water

An insulated beaker with negligible mass contains liquid water with a mass of 0.340 kg and a temperature of 66.3 ∘C .

How much ice at a temperature of -17.9 ∘C must be dropped into the water so that the final temperature of the system will be 22.0 ∘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 fusion for water to be 3.34×10⁵ J/kg .

Answer 1

Heat energy gained by ice is equal to heat enrgy lost by water.
Heat energy gained by ice is sum of energy required to
1) increase it's tempreture from - 17.9 ^oC to 0 ^oC = MC(delta T) = M(2100) ( 0 - (- 17.9) ) = 37590 M (M is mass of ice and C is specific heat of ice)
2) melt it = ML = M (3.34x10⁵) L is latent heat of fusion.
3) increase it's tempreture from 0 to 22^oC = MC(delta T) = M (4190) (22) = 92180 M

Total heat energy gained 463770 M

Heat energy lost by water = MC(delta T) = 0.34(4190) ( 66.3-22) = 63109.78 (M is mass of water and C is specific heat of water)Equating energy lost by water to that gained by ice , we get
63109.78 = 463770M
M = 0.136 kg