ΔQ=8575 cal;ΔQ≈4.2∗8575=36015 J=36 kJ
m=15g=0.015 kg
T=100°+273=373 K
Q1 the amount of heat in the transition from steam to liquid
L=2260kgkJ
Q1=Lm=2260∗0.015=33.9 kJ
Q2=ΔQ−Q1=36−33.9=2.1 kJ
Q2 the amount of heat when the temperature changes
Q2=cmΔT
c=4.2kg∗KkJ
ΔT=cmQ2=4.2∗0.0152.1=33 K
T2=T−ΔT=373−33=340 K=67°
Answer: 67° final temperature
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