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29 August, 10:36

How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC? (The specific heat for water is 4.18 J/g·ºC)

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Answers (2)
  1. 29 August, 10:41
    0
    4235.7J or 4.2KJ

    Explanation:

    The heat required to raise the temperature of water is given by

    H = mcθ

    Where

    H = heat required=?

    m = mass of the water body=65.8 grams

    c = heat capacity of the water = 4.18 J/g·ºC

    θ=temperature change = 46.9°C - 31.5°C = 15.4°C

    H = 65.8 * 4.18 * 15.4 = 4235.7J or 4.2KJ
  2. 29 August, 10:47
    0
    q = (4.18/g' C) * (65.8g) * 4069'C - 31.5'C=4240
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