Question

A cube of ice is taken from the freezer at -9.5 ?Cand placed in a 95-g aluminum calorimeter filled with 330 g of water at room temperature of 20.0 ?C. The final situation is observed to be all water at 17.0 ?C. The specific heat of ice is 2100 J/kg?C?, the specific heat of aluminum is 900 J/kg?C?, the specific heat of water is is 4186 J/kg?C?, the heat of fusion of water is 333 kJ/Kg.

What was the mass of the ice cube?

Answer 1

let mass of the ice cube be m kg.

then heat lost by aluminum + heat lost by water=heat gained by ice

==>mass of aluminum *specific heat of aluminum*decrease in temperature +mass of water*specific heat of water*decrease in temperature=mass of ice *specific heat of ice*temperature increase from -9.5 to 0 degree + mass of ice*latent heat of fusion+mass of melted water*specific heat of water*temperature change from 0 to 17 degrees

==>0.095*900*(20-17)+0.33*4186*(20-17)=m*2100*(0-(-9.5))+m*333*10^3+m*4186*(17-0)

==>4400.6=424112*m

==>m=0.010376 kg

=10.376 grams