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12 January, 10:59

A 3.5 kg gold bar at 94°C is dropped into 0.20 kg of water at 22°C What is the final temperature? Assume the specific heat of gold is 129 J/kg C Answer in units of °C

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  1. 12 January, 13:16
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    47.17 degree C

    Explanation:

    mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C

    mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C

    Let T be the temperature at equilibrium.

    Heat given by the gold = Heat taken by water

    mg x sg x (T1 - T) = mw x sw x (T - T2)

    3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)

    42441 - 451.5 T = 840 T - 18480

    60921 = 1291.5 T

    T = 47.17 degree C
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