The amount of heat gained by a solid object to convert it into a liquid without any further increase in the temperature is known as latent heat of fusion. The content of latent heat is complex in the case of sea ice because it is possible for sea ice and brine to exist together at any temperature and melt at a temperature other than 0oC when bathed in a concentrated salt solution, just like it occurs in the walls of brine cells when brine cells migration occurs. If m kg of solid converts to a fluid at a constant temperature that is its melting point, the heat consumed by the substance or the latent heat of fusion formula is expressed as
Q = m × L
L = specific latent heat of fusion of substance.
The temperature of the substance changes from t1 (low temperature) to t2 (high temperature) the heat which the material absorbs or releases is expressed as
Q = mc Δt
Q = mc (t2 – t1)
The total amount of heat absorbed or liberated by the material is
Q = mL + mc Δt
The latent heat-related concept is marked below in the table
|Latent Heat of Water|
A piece of metal at 20oC has a mass of 60g. When it is immersed in a current of steam at 100∘C, 0.5g of steam is condensed on it. Determine the specific heat of metal, given that the latent heat of steam = 540 cal/g.
Let c be the specific heat of the metal.
Heat gained by the metal
Q = m c Δt
Q = 60 × c × (100 – 20)
Q = 60 × c × 80 cal
The heat released by the steam
Q = m × L
Q = 0.5 × 540 cal
By the principle of mixtures,
Heat given is equal to Heat taken
0.5 × 540 = 60 × c ×80
c = 0.056 cal/g ∘C
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