Latent Heat of Fusion Formula
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 changes to liquid at constant temperature which is its melting point, the heat absorbed by the substance or the latent heat of fusion formula is given by,
Q = m L
L = specific latent heat of fusion of substance.
If the temperature of the object varies from the lower temperature t1 to higher temperature t2, the heat gained or liberated by the material is given by,
Q = mc Δt
= mc (t2 – t1)
Therefore, the total heat gained or released by the substance will be,
Q = mL + mc Δt
Following is the table explaining latent heat related concept
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
= m c Δt
= 60 × c × (100 – 20)
= 60 × c × 80 cal
Heat given by steam = mL = 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.
Calculate the result if 64500 calories of heat are extracted from 100 g of steam at 100∘C. Latent heat of ice and steam are given as 80 cal/g and 540 cal/g respectively.
If the full steam is converted into the water at 100∘C, then the temperature of water falls from 100∘C to 10∘C and then a part of water at 0∘C is turned into ice.
If the water converted into ice = mg
The heat gained to bring the steam at 100∘C to water at 100∘C = mL steam
= 100 × 540 cal
= 54000 cal.
The heat gained to bring water at 100∘C into water at 0∘C
= 100 × 1 × (100-0)
The total heat extracted,
= 54000 + 10000
= 64000 cal.
Heat still remained
= 64500 – 64000
= 500 cal.
Amount of water converted to ice,
m = Q / L
= 500 / 80
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