Answer:
The amount of ice added in gram is 32.77g
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
