Question

How much heat is required to change m=1kg of ice at -6°C into water at 60°C?...

How much heat is required to change m=1kg of ice at -6°C into water at 60°C? cw= 4190 J/kg*K (specific heat of water), Lf= 333 kJ/kg (latent heat of fusion of ice), cice= 2100 J/kg*K

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Answer #1

Ti = -6.0 oC

Tf = 60.0 oC

here

Cs = 2100 J/Kg.oC

Heat required to convert solid from -6.0 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 1 Kg * 2100 J/Kg.oC *(0--6) oC

= 12600 J

Lf = 333KJ/Kg =

333000J/Kg

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 1.0 Kg *333000 J/Kg

= 333000 J

Cl = 4190.0 J/Kg.oC

Heat required to convert liquid from 0.0 oC to 60.0 oC

Q3 = m*Cl*(Tf-Ti)

= 1 Kg * 4190 J/Kg.oC *(60-0) oC

= 251400 J

Total heat required = Q1 + Q2 + Q3

= 12600 J + 333000 J + 251400 J

= 597000 J

= 597 KJ

Answer: 597 KJ

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