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9 October, 20:32

If 24500 J is applied to 125g of water at 35 C, what will the final temperature of the water be?

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  1. 9 October, 20:41
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    The final temperature of water is calculated using the below equation

    Q = MC delta T

    Q = heat energy (24500 j)

    m = mass (125 g)

    c = specific heat capacity (4.18 j/g/c)

    delta T = change in temperature (final temperature - initial temperature) = (T-35)

    final temperature is therefore=24500 j = 125 g x 4.18 g/j/c x (T - 35)

    24500 j=522.5 (T-35 c)

    24500 j = 522.5T - 18287.5

    like terms together 24500 + 18287.5 = 522.5 T

    42787.5j = 522.5 T

    divide both side by 522.5 T

    T=81.9 c
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