Question

Example 11.4 When 0.15 kg of ice at 0 °C
is mixed with 0.30 kg of water at 50°C in a
container, the resulting temperature is
6.7 °C. Calculate the heat of fusion of ice.
(s = 4186 J kg K)
water​

Answer 1

Answer:

334461.4 J/kg

or

334.46 J/g

Explanation:

Initial Temperature of water (t1) = 50°C

mass of water (m1) = 0.30 kg

Initial temperature of ice (t2) = 0°C

mass of ice (m2) = 0.15 kg

final temperature (t)= 6.7°C

Specific heat capacity of water (C) = 4186 J/kg K (or 4186 J/kg°C)

Latent heat of fusion of ice = x/kg

Heat gained = Heat Lost

(latent heat of fusion of ice*mass of ice) +(specific heat of water* mass of ice* change in temperature of ice) = Specific heat of water*mass of water*change in temperature of water

(x*m2)+(4186*m2*(t-t2))=(4186*m1*(t1-t))

(x*0.15)+(4186*0.15*(6.7-0))=(4186*0.30*(50-6.7))

0.15x + 627.9*6.7 = 1255.8 * 43.3

0.15x + 4206.93 = 54376.14

0.15x = 50169.21

x = 50169.21 / 0.15

x = 334461.4 J/kg