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mH₂O= 10kg
T₁=200°C
T₂=1000°C
T₃=300°C-200°C=100°C
T₄=1000°C-300°C=700°C
cwH₂O=4190 J/kg*K
cwFe=440 J/kg*K
Szukane:
mFe=?
Wzory:
Qpobrane=Qoddane
Q=cw*m*T
Rozwiązanie:
cwH₂O*mH₂O*T₃=cwFe*mFe*T₄
mFe=cwH₂O*mH₂O*T₃/cwFe*T₄
mFe=4190J/kg*K×10kg×100°C÷440J/kg*K×700°C≈13,6kg
Odp. Masa żelaza wynosi około 13,6kg