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1 September, 10:40

A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 joules what is the approximate final temperature of the water.

74 C

78C

81 C

83C

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Answers (1)
  1. 1 September, 11:00
    0
    83°C

    Explanation:

    The following were obtained from the question:

    M = 40g

    C = 4.2J/g°C

    T1 = 91°C

    T2 = ?

    Q = 1300J

    Q = MCΔT

    ΔT = Q/CM

    ΔT = 1300 / (4.2x40)

    ΔT = 8°C

    But ΔT = T1 - T2 (since the reaction involves cooling)

    ΔT = T1 - T2

    8 = 91 - T2

    Collect like terms

    8 - 91 = - T2

    - 83 = - T2

    Multiply through by - 1

    T2 = 83°C

    The final temperature is 83°C
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