kawałek stali o masie 2 kg i temperaturze 400 K włożono do wody o temeraturze 5 stopni. Temperatura koncowa wody wynosila 60 stopni. Oblicz mase wody.
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m1=2kg
T1=400K
T2=5C=278K
T3=60C=333K
m2=?
Cw stali=460J/kgK
Cw=4200j/kgK
Q1+Q2=Q3
Q1=m1CsT1
Q2=m2CwT2
Csm1T1+Cwm2T2=CwT3(m1+m2)
Cwm2T2-CwT3(m1+m2)=Csm1T1
Cwm2T2+CwT3m2=Csm1T1+CwT3m1
M2(CwT2+CwT3)=M1(CsT1+CwT3)
m2=(M1(CsT1+CwT3)/(CwT2+CwT3)
m2=(2(460*400+4200*333)/(4200*378+4200*333)
m2=2(1582600)/2986200
m2=1,05kg
odp masa wody to 1,05kg