Question

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)

Answer #1

given

mass m = 0.347 kg

piece of ice from -12.6ºC to 132.1ºC

C_{ice} = 2090 J/kgºC,

C_{water} = 4186 J/kgºC,

C_{steam} = 1520 J/kg,

L_{vap} = 2256 kJ/kg

and L_{fus} = 334 kJ/kg

so the total heat can be written as

Q = the energy needed for -12.6^{o}C
to0^{o}C

+ energy needed for L_{fus}

+ energy needed for 0^{o}C water to 100^{o}C

+ energy needed for 100^{o}C water to 100^{o} C
vapour

+ energy needed for 100^{0}C to 132.1^{o}C

Q = m C_{ice} ( 12.6 ) + m L_{fus} + m
C_{water} X 100 + m L_{vap} + m C_{steam} (
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 10 ^{6} J**

**the energy needed for phase changes is Q =
1.070052922 X 10 ^{6} J**

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