1个回答
展开全部
设平衡温度T体系绝热则Q=ΔfusH+500g*Cp(T-273.15K)+1000g*Cp(T-353.15K)=333.3*500J+1500g*CpT-489725g·K*Cp=0解T=299.93K=26.78℃
水熔化程ΔfusG=ΔfusH-TfusΔfusS=0即ΔfusS=ΔfusH/Tfus=333.3*500J/273.15K=610.1J/K
变温传热程看作数TT±dT平衡传热程累积δQ=CdTdS=δQ/T=CdT/T式作T0T积ΔS=S(T)-S(T0)=C(lnT-lnT0)=Cln(T/T0)
0.5kg0℃水升温至26.78℃C=500g*Cp=2092J/KT0=273.15KT=299.93KΔS1=2092J/Kln(299.93/273.15)=195.66J/K
1kg80℃水降温至26.78℃C=1000g*Cp=4184J/KT0=353.15KΔS2=4184J/Kln(299.93/353.15)=-683.43J/K
所程总熵变ΔS=ΔfusS+ΔS1+ΔS2=(610.1+195.66-683.43)J/K=122.33J/K
水熔化程ΔfusG=ΔfusH-TfusΔfusS=0即ΔfusS=ΔfusH/Tfus=333.3*500J/273.15K=610.1J/K
变温传热程看作数TT±dT平衡传热程累积δQ=CdTdS=δQ/T=CdT/T式作T0T积ΔS=S(T)-S(T0)=C(lnT-lnT0)=Cln(T/T0)
0.5kg0℃水升温至26.78℃C=500g*Cp=2092J/KT0=273.15KT=299.93KΔS1=2092J/Kln(299.93/273.15)=195.66J/K
1kg80℃水降温至26.78℃C=1000g*Cp=4184J/KT0=353.15KΔS2=4184J/Kln(299.93/353.15)=-683.43J/K
所程总熵变ΔS=ΔfusS+ΔS1+ΔS2=(610.1+195.66-683.43)J/K=122.33J/K
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
