When 0.15kg of ice at 0°C is mixed with 0.30kg of water at 50°C in a container, the resulting temperature of the mixture is 6.7°C. Calculate the latent heat of the fusion of ice.
By BYJU'S Exam Prep
Updated on: September 13th, 2023
(Swater = 4186 J kg-1K-1)
(a) 3.34 x 105 J kg-1
(b) 1.5 x 105 J kg-1
(c) 2 x 105 J kg-1
(d) 5.4 x 105 J kg-1
Given, 0.15 kg of ice at 0°C is mixed with 0.30kg of water at 50°C.
The resulting temperature of the mixture is 6.7°C.
We have to find the latent heat of the fusion of ice.
Loss of heat by water = Mwater x Swater x ΔT
Where ΔT = change in temperature
ΔT = 50 – 6.7 = 43.3°C
Now, loss of heat by water = 0.30 x 4186 x 43.3
= 54376.14 J
Heat gained by ice = heat required to melt the ice + heat required to raise the temperature of ice water to the final temperature
= Mice x Lfusion + Mice x Swater x ΔT
Where ΔT = change in temperature
ΔT = 6.7 – 0 = 6.7°C
So, heat gained by ice = 0.15 x Lfusion + 0.15 x 4186 x 6.7
= 0.15 Lfusion + 4206.93 J
We know that heat lost is equal to the heat gained.
So, 54376.14 = 0.15 Lfusion + 4206.93
0.15 Lfusion = 54376.14 – 4206.93
0.15 Lfusion = 50169.21
Lfusion = 50169.21/0.15
Lfusion = 3.34 x 105 J kg-1
Therefore, the required latent heat of fusion of ice is 3.34 x 105 J kg-1.
Summary:
When 0.15kg of ice at 0°C is mixed with 0.30kg of water at 50°C in a container, the resulting temperature of the mixture is 6.7°C. Calculate the latent heat of the fusion of ice.
When 0.15kg of ice at 0°C is mixed with 0.30kg of water at 50°C in a container, the resulting temperature of the mixture is 6.7°C. The latent heat of fusion of ice is 3.34 x 105 J kg-1.
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