4 Latent heat Heat of fusion Heat of vaporization – Problems and Solutions
1. Calculate the amount of heat added to 1 gram gold to change phase from solid to liquid. The heat of fusion for gold is 64.5 x 103 J/kg.
Known :
Mass (m) = 1 gram = 1 x 10-3 kg
Heat of fusion (LF) = 64.5 x 103 J/kg
Wanted : Heat (Q)
Solution :
Q = m LF
Q = (1 x 10-3 kg)(64.5 x 103 J/kg)
Q = 64.5 Joule
2. Calculate the amount of heat released by 1 gram mercury to change phase from liquid to solid. Heat of fusion for mercury is 11.8 x 103 J/kg.
Known :
Mass (m) = 1 gram = 1 x 10-3 kg
Heat of fusion (LF) = 11.8 x 103 J/kg
Wanted : Heat (Q)
Solution :
Q = m LF
Q = (1 x 10-3 kg)(11.8 x 103 J/kg)
Q = 11.8 Joule
3. Determine the amount of heat absorbed by 1 kg water to change phase from liquid to vapor (steam). Heat of vaporization for water = 2256 x 103 J/kg
Known :
Mass (m) = 1 kg
Heat of vaporization (LV) = 2256 x 103 J/kg
Wanted : Heat (Q)
Solution :
Q = m LV
Q = (1 kg)(2256 x 103 J/kg)
Q = 2256 x 103 Joule
4. Determine the amount of heat released by nitrogen to change phase from vapor to liquid. Heat of vaporization for nitrogen = 200 x 103 J/kg
Known :
Mass (m) = 1 gram = 1 x 10-3 kg
Heat of vaporization (LV) = 200 x 103 J/kg
Known : Heat (Q)
Solution :
Q = m LV
Q = (1 x 10-3 kg)(200 x 103 J/kg)
Q = 200 Joule
- Converting temperature scales
- Linear expansion
- Area expansion
- Volume expansion
- Heat
- Mechanical equivalent of heat
- Specific heat and heat capacity
- Latent heat, heat of fusion, heat of vaporization
- Energy conservation for heat transfer