Question

An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equilibrium with a final temperature of 12∘C. How much ice in grams was added to the water?

The specific heat of ice is 2090 J/(kg ∘C), the specific heat of water is 4186 J/(kg ∘C), latent heat of the ice to water transition is 3.33 x10^5 J/kg

Answer 1

The amount of ice added in gram is 32.77g

Step-by-step explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g