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.

Loss of heat by water = M_water x S_water 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

= M_ice x L_fusion + M_ice x S_water x ΔT

Where ΔT = change in temperature

ΔT = 6.7 – 0 = 6.7°C

So, heat gained by ice = 0.15 x L_fusion + 0.15 x 4186 x 6.7

= 0.15 L_fusion + 4206.93 J

We know that heat lost is equal to the heat gained.

So, 54376.14 = 0.15 L_fusion + 4206.93

0.15 L_fusion = 54376.14 – 4206.93

0.15 L_fusion = 50169.21

L_fusion = 50169.21/0.15

L_fusion = 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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