Question

An insulated beaker with negligible mass contains a mass of 0.300 kg of water at a temperature of 63.3
c. How many kilograms of ice at a temperature of 12.1 c must be dropped in the water to make the final temperature of the system 36.9 c?

Answer 1

Heat is lossed by the water and gained by the ice

The sp. Heat capacity of ice is about 2100 J/(kg °k) and water is 4200 ......

While converting ice into water at constant temp. of 0°C the latient heat of fusion Lf= 80 *4200= 336000J/kg

Now lets go to the problem

for water

H1= msΔt= 0.3*4200*(63.3-36.9)

= 33264 J

For ice first it change from -12.1 to 0°C and then from ice to water at 0°C and from 0 to 36.9 °C

So for ice

H2=

m *2100*12.1+m*336000+m*4200*36.9

=m(25410+336000+154980)

=516390m

According to calorimetry

Heat lossed by water = heat gained

by ice

i.e. H1=H2

33264=516390m

m=0.0644 kg