**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 10^{3 }J/kg.

__Known :__

Mass (m) = 1 gram = 1 x 10^{-3} kg

Heat of fusion (L_{F}) = 64.5 x 10^{3 }J/kg

__Wanted :__ Heat (Q)

__Solution :__

Q = m L_{F}

Q = (1 x 10^{-3} kg)(64.5 x 10^{3 }J/kg)

Q = 64.5 Joule

[irp]

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 10^{3 }J/kg.

__Known :__

Mass (m) = 1 gram = 1 x 10^{-3} kg

Heat of fusion (L_{F}) = 11.8 x 10^{3 }J/kg

__Wanted :__ Heat (Q)

__Solution :__

Q = m L_{F}

Q = (1 x 10^{-3} kg)(11.8 x 10^{3 }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 10^{3} J/kg

__Known :__

Mass (m) = 1 kg

Heat of vaporization (L_{V}) = 2256 x 10^{3} J/kg

__Wanted :__ Heat (Q)

__Solution :__

Q = m L_{V}

Q = (1 kg)(2256 x 10^{3} J/kg)

Q = 2256 x 10^{3} Joule

[irp]

4. Determine the amount of heat released by nitrogen to change phase from vapor to liquid. Heat of vaporization for nitrogen = 200 x 10^{3} J/kg

__Known :__

Mass (m) = 1 gram = 1 x 10^{-3} kg

Heat of vaporization (L_{V}) = 200 x 10^{3} J/kg

__Known :__ Heat (Q)

__Solution :__

Q = m L_{V}

Q = (1 x 10^{-3} kg)(200 x 10^{3} 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