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dane: m=300kg, cw=4,2 kJ/kg*C, t1=80*C, t2=60*C
szukane: Q
Q = m*cw*[t1-t2] = 300kg*4,2kJ/kg*C*20*C = 25 200 kJ = 25,2 MJ
Odp.: Woda oddaje 25,2 MJ ciepła w ciągu doby w zadanych warunkach.
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