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13 March, 16:05

What is the final temperature of 78g sample of water at 99°C if it loses 6500 J of heat energy?

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  1. 13 March, 19:25
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    The specific heat of water is 4.184 J/g/°C,

    Therefore: 78 g of water at 99°C loses 78 (4.184) = 326 J per °C.

    (6500) J / (326 J/°C) = 19.93 °C

    So the final temperature is 99-19.93 = 79.07 °C
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