Question

A calorimeter of mass 50 g and specific heat capacity 0.40Jg'°C-contains 200 g of water
at 40°C. What amount of ice at 0°C should be added to it so that the final temperature is
10°C? [specific heat capacity of water = 4.2 Jg !°C- and specific latent heat of fusion of
Te = 336] g-'°C-
​

Answer 1

Explanation:

Heat energy given by metal piece =m⋅c⋅ΔT

1

=20×0.3×(100−22)

=468J

Heat energy gained by water =m

w

×c

w

×ΔT

2

=m

w

×4.2×(22−20)

=m

w

×8.4J

Heat energy gained by calorimeter =m

c

×c

c

×ΔT

2

=50×0.42×(22−20)

=42J

By principle of calorimeter:

Heat lost = Heat gained

Heat energy given by metal = Heat energy gained by water + Heat energy gained by calorimeter

468=m

w

×8.4+42

m

w

=

8.4

468−42

=50.7g

Therefore, the mass f water used in the calorimeter is 50.7g.