Latent Heat of Fusion Formula with Sample Problems

Latent Heat of Fusion Formula :

Latent heat of fusion Q(J/g) In joule per gram is equal to the product of the  mass of body m(kg) in kilogram is multiply by specific latent heat of fusion L(J/g) joule per gram in the addition of mass of body and the specific latent heat of fusion  c (J)  in  joule and the temperature change during heat absorption Δt(J).Hence the Latent Heat Of Fusion Formula has been written as,

Q(J/g)  = m(kg) L(J/g) + m(kg) c (J) Δt(J)

where,

Q→ latent heat of fusion J/g.

m→ mass of body in kg.

L →specific latent heat of fusion in J/g.

c →specific heat of body in J.

Δt→ temperature change during heat absorption (W).

Sample Problems :

Example 1:

Calculate the specific heat of metal, given that the latent heat of steam = 550 cal/g? In a piece of metal at 30oC has a mass of 70g. When it is immersed in a current of steam at 100C, 0.6g of steam is condensed on it.

Solution :

Let c be the specific heat of the metal.

Heat gained by the metal

Q = m c Δt

Q = 70 × c × (100 – 30)

Q = 70 × c × 70 cal

The heat released by the steam

Q = m × L

Q = 0.6 × 550 cal

By the principle of mixtures,

Heat given is equal to Heat taken

0.6 × 550 = 70 × c ×70

c = 1.484 cal/g C .

Example 2

Calculate the specific heat of metal, given that the latent heat of steam = 580 cal/g? In a piece of metal at 40oC has a mass of 60g. When it is immersed in a current of steam at 100C, 0.7g of steam is condensed on it.

Solution:

Let c be the specific heat of the metal.

Heat gained by the metal

Q = m c Δt

Q = 60 × c × (100 – 40)

Q = 60 × c × 60 cal

The heat released by the steam

Q = m × L

Q = 0.7 × 580 cal

By the principle of mixtures,

Heat given is equal to Heat taken

0.7 × 580 = 60 × c ×60

c = 8.8669 cal/g C .

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