Answer to Question #321048 in Molecular Physics | Thermodynamics for Nichole

Question

4kg of water at 50°C is poured on a large block of ice at 0°C. How much I've melts. Specific heat capacity of water= 4200J/Kg/°C. Latent heat of fusion of ice= 3.33x10^5J

Answer 1

Given:

m₁(H₂O)=4 kg

T=50°C

T(ice)=0°C

Specific heat capacity of water; S= 4200J/Kg/°C

Latent heat of fusion of ice; Lf= 3,33x10^5J

m₂(ice melted)=?

Solution:

Heat flow from higger temperature body to low temperature body

Heat lost by water=Heat gained by ice

Heat lost by water=m*S*ΔT=4*4200*50=840000 J

Heat gained by ice=m*Lf

m₁*S*ΔT=m₂*Lf

m₂=m₁*S*ΔT/Lf

m₂=4*4200*50/3.33*10⁵=2,52kg

Answer: 2,52 kg of ice melted