Problem: If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?

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Assuming water is at room temperature (25°C) at the beginning. Specific heat of water is 4.18 J/g°C

Q = mCT(14.5 kJ)(1x103 J1 kJ) =(485 g)(4.18 Jg°C)(Tf - 25°C)Tf - 25°C = (14.5 kJ)(1x103 J1 kJ) (485 g)(4.18 Jg°C)Tf -25°C = 7.15 °C

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If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?

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