Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.250 kg and a temperature of 62.0 ∘C .

How much ice at a temperature of -25.0 ∘C must be dropped into the water so that the final temperature of the system will be 39.0 ∘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

let ice required be m kg.

heat lost by water=mass*specific heat of water*temperature difference

=0.25*4190*(62-39)=24092.5 J

this much heat will be used to first raise the temperature of m kg ice from -25 degree celcius to 0 degree celcius,

melt m kg of ice and then raise the temperature of m kg of liquid water from 0 degree celcius to 39 degree celcius

heat required to raise the temperature of m kg of ice from -25 degree celcius to 0 degree celcius=mass*specific heat of ice*temperature difference=m*2100*(0-(-25))=52500*m J

heat required to melt m kg of ice=mass*latent heat of fusion=334000*m J

heat required to raise the temperature of m kg liquid water from 0 degree celcius to 39 degree celcius=mass*specific heat of water*temperature difference=m*4190*(39-0)=163410*m J

total heat energy required=52000*m+334000*m+163410*m=549410*m J

as there has been no heat loss to the surrounding,

heat lost by water=heat gained by ice

==>24092.5=549410*m

==>m=0.04385 kg

hence mass of ice required is 0.04385 kg.