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3 September, 06:47

The temperature of a sample of water increases from 30.0 degrees C to 40.0 degrees C as 6540 joules of heat are added. What is the mass of the sample of water?

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Answers (2)
  1. 3 September, 06:55
    0
    166 g

    Explanation:

    The equation of specific heat capacity is:

    ΔE = mcΔtheta

    ΔE = 6540J

    c = 4200J/kg°C

    Δtheta = 40 - 30 = 10

    6540 = m * 4200 * 10

    6540 = m * 42000

    6540 : 42000 = m

    m = 0.15571428 ...

    m = 0.166 kg

    m = 166 g
  2. 3 September, 07:06
    0
    Answer: 166 g

    Explanation: q = mC∆T

    q = heat = 6940 J

    m = mass = ?

    C = specific heat of water = 4.184 J/g/deg

    ∆T = change in temperature = 40.0 - 30.0 = 10.0ºC

    6940 J = (m) (4.184J/g/deg) (10 deg)

    m = 165.9 g = 166 g
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