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leraaeezh
@leraaeezh
July 2022
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сколько килограммов 100 градусного пара потребуется для нагревания 80 литров воды от 6 градусов до 35.
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pokrovskaya01
Q1=Q2
Q1=cm1(t-t1)
Q2=Lm+cm(t2-t); m1=80кг; L=2260000 Дж/кг; с=4200 Дж/(кг*град) , t=35; t1=6; t2=100
m=cm1(t-t1)/(L+c(t2-t))
m=4200*80*29/(2260000+4200*65)=3.85 кг
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Answers & Comments
Q1=cm1(t-t1)
Q2=Lm+cm(t2-t); m1=80кг; L=2260000 Дж/кг; с=4200 Дж/(кг*град) , t=35; t1=6; t2=100
m=cm1(t-t1)/(L+c(t2-t))
m=4200*80*29/(2260000+4200*65)=3.85 кг