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
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 !
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