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 Shivank Goel|Updated : August 26th, 2022

(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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