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16 July, 06:19

You pour 170 g hot coffee at 78.7°c and some cold cream at 7.50°c to a 115-g cup that is initially at a temperature of 22.0°c. the cup, coffee, and cream reach an equilibrium temperature of 58.0°c. the material of the cup has a specific heat of 1091 j / (kg · k) and the specific heat of both the coffee and cream is 4190 j / (kg · k). assume that no heat is lost to the surroundings or gained from the surroundings. (a) does the cream lose heat or gain heat?

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  1. 16 July, 09:56
    0
    Because all temperatures are below boiling temperature, only sensible heat exchange occurs.

    The equilibrium temperature is 58 °C.

    Coffee:

    initial temperature = 78.7 °C

    mass = 170 g = 170 x 10⁻³ kg

    specific heat = 4190 J / (kg-K)

    temperature drop = 78.7 - 58 = 20.7 K

    Because the tempeature drops, the coffee loses heat.

    Heat loss = (170x10⁻³ kg) * (4190 J / (kg-K) * (20.7 K)

    = 14744.6 J

    = 14.745 kJ

    Cream:

    initial temperature = 7.5 °C

    Assume that the mass is m kg (not given)

    specific heat (same as for coffee)

    temperature rise = 58 - 7.5 = 50.5 K

    Because the temperature rises, the cream gains heat.

    Heat gained = (m kg) * (4190 J / (kg-K) * (50.5 K)

    = 211595m J

    = 211.595m kJ

    Cup:

    initial temperature = 22 °C

    mass = 115 g = 115 x 10⁻³ kg

    specific heat = 1091 J / (kg-K)

    temperature rise = 58 - 22 = 36 K

    Because of temperature rise, the cup gains heat.

    Heat gained = (115x10⁻³ kg) * (1091 J / (kg-K)) * (36 K)

    = 4529.2 J

    = 4.53 kJ

    Assume no heat is lost to the surroundings. For energy balance (with heat gain as positive and heat loss as negative), obtain

    -14.475 + 211.595m + 4.53 = 0

    m = 0.047 kg = 47 g

    Answer:

    47 g of cream were added, and the cream gained heat.
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