Question

Un corp de fier cu masa m=500g si temperatura tFe=800°C este introdus in 10 l de apa cu masamH2O=10kg cu temperatura tH2O=20°C.Calculati temperatura finala/temp. de echilibru.

Answer 1

m₁=500 g=0,5 kg

c₁=459,8 J/kg*grad

Qced=Qprim

m₁c₁(t₁-θ)=m₂c₂(θ-t₂)<=>0,5*459,8(800-θ)=10*4185(θ-20)<=>229,5(800-θ)=41850(θ-20)<=>800-θ=182,35(θ-20)<=>800-θ=182,35θ-3647<=>-183,35θ=-4447=>θ=4447/183,35=24,25°C

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