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18 November, 11:16

The burning of a sample of propane

generated 104.6 kJ of heat. All of this

heat was use to heat 500.0 g of water

that had an initial temperature of

20.0/C. What was the final temperature

of the water? (70.0/C)

+2
Answers (1)
  1. 18 November, 14:47
    0
    70.0°C

    Explanation:

    We are given;

    Amount of heat generated by propane as 104.6 kJ or 104600 Joules Mass of water is 500 g Initial temperature as 20.0 ° C

    We are required to determine the final temperature of water;

    Taking the initial temperature is x°C

    We know that the specific heat of water is 4.18 J/g°C

    Quantity of heat = Mass * specific heat * change in temperature

    In this case;

    Change in temp = (x-20) ° C

    Therefore;

    104600 J = 500 g * 4.18 J/g°C * (x-20)

    104600 J = 2090x - 41800

    146400 = 2090 x

    x = 70.0479

    =70.0 °C

    Thus, the final temperature of water is 70.0°C
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