16.25g Of Water At 54.0c Releases 402.7J. What Will Be Its Final Temp?
we are working with temperature and i do not understand it can you help me with this problem?
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We were trying to get an answer to this question. This is my daughters homework, and we need help with the answer. Thanks!
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Presumably, the warm water is releasing the heat to some cooler surface or fluid. A joule equates to 0.239 calories. A calorie is the amount of heat it would take to raise one gram of water one degree C. Therefore (402.7 joules x 0.239 calories/joule) would be 96.2 calories. If 16.25 grams of water released that much heat the temperature would be depressed by (96.2/16.25) = 5.92 degrees C. Therefore, the 54 degree C water would reach a temperature of (54-5.9) = 48.1 degrees C.
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