Question

You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K. The enthalpy of fusion of ice at 0 °C is 333 J/g.) Mass of ice = g

Answer 1

`m_(ice) = 65.336\,g`

Step-by-step explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

`Q_(water) = Q_(ice)`

`(100\,g)\cdot \left(4.184\,(J)/(kg\cdot ^(\textdegree)C)\right)\cdot (52\,^(\textdegree)C - 0\,^(\textdegree)C) = m_(ice)\cdot \left(333\,(J)/(g) \right)`

The amount of ice that is melt is:

`m_(ice) = 65.336\,g`