# 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

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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

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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

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