An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 degrees Celsius and rejects h

Question

An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 degrees Celsius and rejects heat to a room at a temperature of 20.6 degrees Celsius. Suppose that liquid water with a mass of 82.1 kg at 0.0 degrees Celsius is converted to ice at the same temperature. Take the heat of fusion for water to be Lf=3.34×105 J/kg.
A. How much heat |QH| is rejected to the room?
B. How much energy E must be supplied to the device?

in progress 0
Huyền Thanh 5 months 2021-08-04T06:51:08+00:00 1 Answers 7 views 0

Answers ( )

    0
    2021-08-04T06:52:43+00:00

    Answer:

    A. Q = 2.74 x 10⁷ J = 27.4 MJ

    B. E = 3.91 x 10⁸ J = 391 MJ

    Explanation:

    A.

    Heat rejected can be found as follows:

    Q = mL

    where,

    Q = Heat rejected = ?

    m = mass = 82.1 kg

    L = Latent Heat of fusion = 334000 J/kg

    Therefore,

    Q = (82.1\ kg)(334000\ J/kg)

    Q = 2.74 x 10⁷ J = 27.4 MJ

    B.

    First, we will calculate the efficiency of the Carnot Cycle as follows:

    Efficiency = 1-\frac{T_1}{T_2}

    where,

    T₁ = Heat intake temperature = 0°C + 273 = 273 k

    T₂ = Heat rejection temperature = 20.6°C + 273 = 293.6 k

    Therefore,

    Efficiency = 1 - \frac{273\ k}{293.6\ k} \\\\Effciency = 0.07

    Therefore, the energy input required is:

    Efficiency = \frac{Q}{E}\\\\E = \frac{Q}{Efficiency}  = \frac{2.74\ x\ 10^7\ J}{0.07}

    E = 3.91 x 10⁸ J = 391 MJ

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )