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20 April, 02:58

A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? What percentage of the heat is used to raise the temperature of (b) the panand (c) the water?

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  1. 20 April, 06:12
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    heat used to rise temperature pan = 30.1%

    heat used to rise temperature water = 69.9%

    Explanation:

    Given data

    mass of water = 0.250 liter = 0.250 kg

    aluminum pan mass = 0.500 kg

    initial temperature = 20.0ºC

    final temperature = 80.0ºC

    to find out

    heat used to rise temperature of pan and water

    solution

    we find here heat transferred to the water that is

    heat transferred to the water = mass of water * specific heat of water * change in temperature ... 1

    specific heat of water is 4186 J/kgºC

    so

    heat transferred to the water = 0.250 * 4186 * (80-20) kJ

    heat transferred to the water = 62.8 kJ

    and

    heat transferred to the aluminum that is

    heat transferred to the aluminum = mass of aluminum * specific heat of aluminum * change in temperature ... 2

    here specific heat of aluminum is 900 J/kgºC

    heat transferred to the aluminum = 0.500 * 900 * (80-20) kJ

    heat transferred to the aluminum = 27 kJ

    so

    total heat = 62.8 + 27 = 89.8 kJ

    so

    heat used to rise temperature pan = 27/89.8 * 100% = 30.1%

    heat used to rise temperature water = 62.8 / 89.8 * 100% = 69.9%
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