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8 January, 02:24

A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what is the final temperature of both substances at thermal equilibrium?

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  1. 8 January, 02:29
    0
    Given:

    Ma = 31.1 g, the mass of gold

    Ta = 69.3 °C, the initial temperature of gold

    Mw = 64.2 g, the mass of water

    Tw = 27.8 °C, the initial temperature of water

    Because the container is insulated, no heat is lost to the surroundings.

    Let T °C be the final temperature.

    From tables, obtain

    Ca = 0.129 J / (g-°C), the specific heat of gold

    Cw = 4.18 J / (g-°C), the specific heat of water

    At equilibrium, heat lost by the gold - heat gained by the water.

    Heat lost by the gold is

    Qa = Ma*Ca * (T - Ta)

    = (31.1 g) * (0.129 J / (g-°C) ( * (69.3 - T °C) -

    = 4.0119 (69.3 - T) j

    Heat gained by the water is

    Qw = Mw*Cw * (T-Tw)

    = (64.2 g) * (4.18 J / (g-°C)) * (T - 27.8 °C)

    = 268.356 (T - 27.8)

    Equate Qa and Qw.

    268.356 (T - 27.8) = 4.0119 (69.3 - T)

    272.3679T = 7738.32

    T = 28.41 °C

    Answer: 28.4 °C
  2. 8 January, 02:41
    0
    The final temperature of both substances at thermal equilibrium is 28.4 degrees Celsius.
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