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A blacksmith heat an iron bar to 1445 Celsius. The blacksmith then tempered the metal by dropping into 42,800 cm^3 of water that had a temperature of 22Celsius. The final temperature of the system was 45 Celsius. What was the mass of the bar?

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so the temperature of the water has risen by \(45-22 = 23^{\circ}C\)

that's the delta t right?

Yes, then the heat added to the water is given by,
\(Q=c*m*\Delta t=4.18*42800*23 = 4114792 J\)

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