Answer to Question #74780 in General Chemistry for Karen

Question

Calculate the energy needed to heat 13.6 g ice at -20.0 °C to liquid water at 75.0 °C. The heat of vaporization of water = 2257 J/g, the heat of fusion of water = 334 J/g, the specific heat capacity of water = 4.18 J/g·°C, and the specific heat capacity of ice = 2.06 J/g·°C.

Answer 1

Solution
The process includes 3 steps: heating ice from -20.0 °C to 0°C; melting ice and heating water
from 0°C to 75.0 °C.
Q = CicemΔT1 + Hfusionm + Cwaterm ΔT2 = m × (CiceΔT1 + Hfusion + CwaterΔT2)
Q = 13.6 × (2.06 × 20 + 334 + 4.18 × 75) = 9366 (J)
Answer
9366 J – the energy needed to heat 13.6 g ice at -20.0 °C to liquid water at 75.0 °C.