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4.50 grams of water at 25.0 °C was heated by the addition of 133 J of energy. What was the final temperature of the water? ThNEED ANSWER ASAP

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Answer #1

m= 4.50 grams c=4.185/9.00 Ti=25.0°C Tf = ? q= 1333 q=MCAT 133 = 4.50 X4.18 (TF-25.0) Tf = 32.1°C

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