Question

When 15.0 g of ice at 0°C is melted in 100 g of water at 20.0°C, the final temperature is 9.5°C. Find the heat of fusion of ice, based on this data.

Answer 1

Here we have given that,

Mass of ice = 15.0 g

Temp of ice = 0°C

Mass of water = 100 g

Temp of water = 20.0°C,

final temperature= 9.5°C.

Specific heat of water s = 4.187 J/g-C

To Find the heat of fusion of ice,

So that

On applying the formula

q = ms(deltaT)

q = 100 × 4.187 × (9.5 -20)

q = -4396.35 J

This represents the heat removed from the water, hence its negative sign.

By the laws of thermodynamics, this means that the ice cubes in the water absorbed +4396.35 J of heat.

So that, To find the heat of fusion of ice, f, according to f = q ÷ m by dividing the heat, q, absorbed by the ice.

f = q/m = 4396.35/15 = 293.09 J/g

Which is our required answer.