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26 November, 08:29

What would the expected temperature change be (in Fahrenheit) if a 0.5 gram sample of water released 50.1 J of heat energy? The specific heat of liquid water is 4.184 J/g-C

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  1. 26 November, 12:23
    0
    75.11°F

    Explanation:

    Given dа ta:

    Change in temperature = ?

    Mass of water = 0.5 g

    Heat released = 50.1 J

    Specific heat of water = 4.184 j/g. °C

    Solution:

    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

    Q = m. c. ΔT

    50.1 J = 0.5 g. 4.184 j/g. °C. ΔT

    50.1 J = 2.092 j / °C.ΔT

    50.1 J / 2.092 j / °C = ΔT

    23.95 °C = ΔT

    °C to °F

    (23.95°C * 9/5) + 32 = 75.11°F
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