the sum of heats gained =5.42e3 kj
heat to warm ice+heat to melt ice+heat to warm water=5.42e3
m*cice*29.7+ m*Hfice + m*cw*(21.7)=5.42e3
solve for mass m, you can look up specific heat of ice cw. You are given Hfice, and cw specific heat water.
How many grams of ice at -29.7 ∘C can be completely converted to liquid at 21.7 ∘C if the available heat for this process is 5.42×103 kJ ?
For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus=6.01kJ/mol .
1 answer
2 answers