Question

The specific heat of ice is 2.100 kJ/kg °C, the heat of fusion for ice at 0°C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, and the heat of vaporization of water at 100.0°C is 2,256 kJ/kg. What is the final equilibrium temperature when 10.00 grams of ice at -15°C is mixed with 3.000 grams of steam at 100°C?

Answer 1

Explanation:

heat gain=heat loss

heat gain of ice=heat loss of steam

heat gain:

mCp dT 1= heat from -15 to 0 deg

mh 1=latent heat of fusion

mCp dT 2= heat from 0 to final temp deg

heat loss:

mCp dT 3=heat from 120 to 100 deg

mh 2= latent heat of vaporization

mCp dT 4= heat from 100 to final temp deg

i assume that the steam will be condensed here

0.02 kg (2.100 kJ/kg °C) (0--15) + 0.02 kg (333.7 kJ/kg) + 0.02 kg (4.186 kJ/kg °C) (Tf-0) = 0.005 (2.020 kJ/kg °C) (120-100) + 0.005 (2,256 kJ/kg) + 0.005 (4.186 kJ/kg °C) (100-Tf)

Tf= 59.16 deg C