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dane m1=0,8kg, t1=20*C, t2=100*C, tk=24*C, co=128J/kgK, cw=4190J/kgK
szukane m2
ciepło oddane = ciepło pobrane
Qo = Qp
m2*co*[t2-tk] = m1*cw*[tk-t1]
m2 = m1*cw*[tk-t1] / co*[t2-tk] = 0,8kg*4190J/kgK*4K / [128J/kgk*76K]
m2 = 13408J / 9728J/kg = 1,378 kg ołowiu
..........................pozdrawiam