Question

How much heat transfer in kJ is necessary to raise the temperature of a 0.347 kg...

How much heat transfer in kJ is necessary to raise the temperature of a 0.347 kg piece of ice from -12.6ºC to 132.1ºC , including the energy needed for phase changes? (cice = 2090 J/kgºC, cwater = 4186 J/kgºC, csteam = 1520 J/kg, Lvap = 2256 kJ/kg, and Lfus = 334 kJ/kg)

Homework Answers

Answer #1

given

mass m = 0.347 kg

piece of ice from -12.6ºC to 132.1ºC

Cice = 2090 J/kgºC,

Cwater = 4186 J/kgºC,

Csteam = 1520 J/kg,

Lvap = 2256 kJ/kg

and Lfus = 334 kJ/kg

so the total heat can be written as

Q = the energy needed for -12.6oC to0oC

+ energy needed for Lfus

+ energy needed for 0oC water to 100oC

+ energy needed for 100oC water to 100o C vapour

+ energy needed for 1000C to 132.1oC

Q = m Cice ( 12.6 ) + m Lfus + m Cwater X 100 + m Lvap + m Csteam ( 132.1 - 100 )

Q = 0.347 ( 2090 X 12.6 + 334000 + 4186 X 100 + 2256000 + 1520 X 32.1 )

Q = 0.347 ( 26334 + 334000 + 418600 + 2256000 + 48792 )

Q = 0.347 X 3083726

Q = 1070052.922 J

Q = 1.070052922 X 106 J

the energy needed for phase changes is Q = 1.070052922 X 106 J

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