Question

. When 50 gram of ice at 0 0C is mixed with 300 gram of water at 50 0C, the resulting temperature is
6.7 0C Calculate the heat of fusion of ice?

Answer 1

To calculate the heat of fusion of ice, we can use the principle of conservation of energy.
The amount of heat lost by the water as it cools down from 50°C to 6.7°C is equal to the amount of heat gained by the ice to melt it at 0°C.
Let Q1 = heat lost by water
Q2 = heat gained by ice
The heat lost by the water can be calculated using the specific heat capacity formula:
Q1 = mcΔT
Where:
m = mass of water (300g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (50°C - 6.7°C)
Q1 = (300g)(4.18 J/g°C)(50°C - 6.7°C)
Q1 = 11124 J
The heat gained by the ice can be calculated using the heat of fusion formula:
Q2 = mLf
Where:
m = mass of ice (50g)
Lf = heat of fusion of ice
Q2 = (50g)(Lf)
Since energy is conserved:
Q1 = Q2
11124 J = 50g * Lf
Lf = 11124 J / 50g
Lf = 222.48 J/g
Therefore, the heat of fusion of ice is 222.48 J/g.