To melt 446 g of ice and heat it to 65 C requires heat transfer to that ice of 446(333 + 65*4186)
Set that number equal to the heat transferred from the steam mass M, which is
M(2256 + 35*4186)
Solve for the single unknown,M, in grams
What mass (in grams) of steam at 100°C must be mixed with 446 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 65.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.
3 answers
816Kg
NO