Physics, asked by saivivek16, 11 months ago

when 0.15 kg of ice of 0 c mixed with 0.30 kg of water at 50 c in a container ,the resulting temperature is 6.7c .Calculate the heat of fusionbof ice.Where ,(s -water=4186 J / kg K ).​

Answers

Answered by Adihya
5

Answer:

Heat lost by water = ms of water

=(0.30kg)(4186 J/Kg K )(50°C -6.7°C)

=54376 .14 J

Heat required to melt ice = m²L^f

= (0.15kg)L^f

Heat required to raise temperature of ice water to final temperature =m¹s^w(theta of f - theta of i ) I

= (0.15 kg)(4186J/Kg K )(6.7°C - 0°C )

=4206.93 J

Heat lost = Heat gained

54376.14 J = (0.15 kg)L^f +4206.93 J

L ^f = 3.34×10^5 J / kg

Thanking you !

Answered by CUTESTAR11
3

Explanation:

In this question we use the concept of specific heat capacity to obtain the latent heat of fusion of ice.

Heat gained by ice = heat lost by the water.

The change in temperature for the water is:

50 - 6.7 =43.3°C

The change in temperature of ice is:

6.7°C

Q=mcΔT where Q is the quantity of heat gained or lost, m the mass of the substance, c the specific heat capacity of water and ΔT the temperature change.

Heat gained by ice is: mLf + mcΔT where lf is the latent heat of fusion and c the specific heat capacity of water.

In this question we are looking for the Lf thus it is not known.

Substituting what we is given in the formulae we get :

0.15Lf + 0.15 × 4186 × 6.7 =0.15Lf + 4206.93

Heat lost by the water is:

0.30 ×4186 × 43.3 =54376.14Joules

54376.14 = 0.15Lf + 4206.93

0.15Lf = 54376.14 - 4206.93

0.15Lf = 50169.21

Lf =50169.21 ÷ 0.15

=334461.40joules/Kg

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