Physics, asked by okechukwucaroline77, 4 months ago

A forty volts electric heater is used to supply a current 12amp for 1400sec to a body of mass 1.5kg at the melting point of the body, the body melts and it's temperature rising through 60degree Celsius in an extra 72secs determine the latent heat of fusion of the body​

Answers

Answered by jahanshu
2

Explanation:

A forty volts electric heater is used to supply a current 12amp for 1400sec to a body of mass 1.5kg at the melting point of the body, the body melts and it's temperature rising through 60degree Celsius in an extra 72secs determine the latent heat of fusion of the body

Answered by soniatiwari214
0

Concept:

The latent heat of fusion is the quantity of energy that must be given to a solid substance (often in the form of heat) to produce a change in its physical state and convert it into a liquid state. It is mathematically expressed as H = Ivt.

Given:

Current, I = 12A

time = 1400 sec

Volts = 40V

mass = 1.5 kg

Extra time = 72 secs

θ = 60⁰C

Find:

We need to determine the latent heat of fusion of the body.

Solution:

The amount of energy that must be added to a solid substance (usually in the form of heat) to cause a change in its physical state and transform it into a liquid is known as the latent heat of fusion, also known as the enthalpy of fusion (when the pressure of the environment is kept constant).

The solid mass was melted at 0⁰C in 1400 sec.

We know, the latent heat of fusion, H = ml where m = mass, l = specific latent heat of fusion of substance.

It has been observed that ml = Ivt

Therefore, formula for latent heat of fusion becomes, H = Ivt

H = 12 × 40 × 1400

H = 672,000 J

It has been given that the melted liquid is heated for another 72 secs through 60⁰C.

We know, H = mcθ

Heat supplied = heat gained

Therefore, the equation becomes, Ivt = mcθ since Ivt = H

12 × 40 × 72 = 1.5 × c × (60-0)

34,560 = 90c

Therefore, c = 384 J/Kg/K

Thus, the latent heat of fusion and specific heat capacity of the body is 672,000J and 384 J/Kg/K, respectively.

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