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20 May, 14:47

If 420 joules of heat energy is added to 25 grams of water at 25 degrees celcius, what will be the final temperature of the water, in Celcius degrees?

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  1. 20 May, 16:56
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    T2 = 29°C

    Explanation:

    Given dа ta:

    Heat added = 420 j

    Mass of water = 25 g

    Initial temperature = 25°C

    Final temperature = ?

    Solution;

    Specific heat capacity:

    It is the amount of heat required to raise the temperature of one gram of substance by one degree.

    Specific heat capacity of water = 4.18 j/g.°C

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    Now we will put the values.

    420 j = 25 g * 4.18 j/g.°C * (Final temperature - initial temperature)

    420 j = 25 g * 4.18 j/g.°C * (T2 - 25°C)

    420 j = 104.5 j/°C * (T2 - 25°C)

    420 j / 104.5 j/°C = T2 - 25°C

    4°C + 25°C = T2

    T2 = 29°C
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