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1 April, 20:53

A 214 g aluminum cup holds and is in thermal equilibrium with 892 g of water at 76°C. The combination of cup and water is cooled uniformly so that the temperature decreases by 0.9°C per minute. At what rate is energy being removed?

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  1. 1 April, 21:33
    0
    The rate at which energy is removed = 59.085 J

    Explanation:

    The rate at which energy is removed = c₁m₁ΔΘ/t + c₂m₂ΔΘ/t ... Equation 1

    Where c₁ = specific heat capacity of the aluminum, m₁ = mass of the aluminum, ΔΘ/t = rate of temperature decrease, c₂ = specific heat capacity of water, m₂ = mass of water.

    Given: m₁ = 214 g = (214/1000) kg = 0.214 kg, m₂ = 892 g = (892/1000) kg

    m₂ = 0.892 kg, ΔΘ/t = 0.9 °C/min. = (0.9/60) °C/seconds. = 0.015 °C/seconds.

    Constants: c₁ = 900 J/kg.°C, c₂ = 4200 J/kg.°C

    Substituting these values into Equation 1,

    The rate at which Energy is removed = (0.214*900*0.015) + (0.892*4200*0.015)

    The rate at which energy is removed = 2.889 + 56.196

    The rate at which energy is removed = 59.085 J
  2. 2 April, 00:11
    0
    Answer: rate at which energy is being removed is 58.9J/s or 58.9W

    Explanation:

    Given;

    Mass of water Mw = 892g = 0.892kg

    Mass of aluminium Ma = 214g = 0.214kg

    Rate of change in temperature ∆T = 0.9°C per minute

    Specific heat capacity of water Cw = 4186J/kgC

    Specific heat capacity of aluminium Ca = 900J/kgC

    Since both the water and aluminium are at thermal equilibrium and the decrease with the same rate, the energy removed per minute is given as;

    Q = (MwCw + MaCa) ∆T

    Q = (0.892*4186 + 0.214*900) * 0.9

    Q = 3533.8608J per minute

    To determine the energy being removed in watts

    P = Q/t

    P = 3533.8608J/60s

    P = 58.89768W

    P = 58.9W
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