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Anees Chandra Chand
Anees Chandra Chand
Asked: 3 years ago2022-11-06T20:37:20+05:30 2022-11-06T20:37:20+05:30In: General Awareness

In a refrigerator one removes heat from a lower temperature and deposits to the surrounding at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.

In a refrigerator one removes heat from a lower temperature and deposits to the surrounding at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.

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  1. 62ed8
    2022-11-01T13:08:22+05:30Added an answer about 3 years ago

    Given :

    T1 = 27°C = 27 + 273 = 300 K

    T2 = −3°C = −3 + 273 = 270 K

    Efficiency,

    η = 1 – \(\frac{T_2}{T_1}\)

    = 1 – \(\frac{270}{300}\)

    = 1 – 0.9

    = 0.1 = \(\frac{1}{10}\)

    Efficiency of refrigerator is 50% of percentage engine.

    η’ = 50% of η = 0.5 ×\(\frac{1}{10}\)=\(\frac{1}{20}\)

    ∴ Coefficient of performance,

    \(β=\frac{Q_2}{W}=\frac{1-η’}{η’}\)

    \(β=\frac{1-\frac{1}{20}}{\frac{1}{20}}=\frac{1-0.05}{0.05}\)

    = \(\frac{0.95}{0.05}\) = 19

    Q2 = 19% of work done by motor on refrigerator

    = 19 × 1

    = 19 K J/S

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Madhu Rajagopal
Madhu Rajagopal
Asked: 3 years ago2022-11-03T16:10:26+05:30 2022-11-03T16:10:26+05:30In: General Awareness

In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from –3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from –3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

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  1. 985e9
    2022-10-29T20:01:19+05:30Added an answer about 3 years ago

    n = 1 – 270/300 = 1/10

    Efficiency of refrigerator = 0.5n = 1/20

    If Q is the heat/s transferred at higher temperture then W/Q = 1/20

    or Q = 20W = 20kJ,

    and heat removed from lower temperture = 19 kJ.

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Rimi Chadha
Rimi Chadha
Asked: 3 years ago2022-10-30T01:01:36+05:30 2022-10-30T01:01:36+05:30In: General Awareness

In a refrigerator, one removes heat from a lower temperature and deposites to the surrounding at a higher temperature. In this process, machanical work has to be done, which is provided by an elecrtic motor. If the motor is of `1KV` power, and heat is transferred from `-3^(@)C “to” 27^(@)C`, find the heat taken out of the refrigerator per second assuming its efficiency is `50%` of a perfect engine.

In a refrigerator, one removes heat from a lower temperature and deposites to the surrounding at a higher temperature. In this process, machanical work has to be done, which is provided by an elecrtic motor. If the motor is of `1KV` power, and heat is transferred from `-3^(@)C “to” 27^(@)C`, find the heat taken out of the refrigerator per second assuming its efficiency is `50%` of a perfect engine.
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  1. 145cd
    2022-11-07T04:28:40+05:30Added an answer about 3 years ago

    Here, power of motor, `W= 1kW`
    `T_(1)= 27^(@)C= (27+273)K= 300 K, T_(2)= -3^(@)C= (-3+273)K= 270K`
    `eta = 1-(T_(2))/(T_(1))= 1-(270)/(300)= 1/(10)`.
    Efficiency of refrigerator `= 0.5 eta= 1/(20)`
    `COP= beta=(Q_(2))/W= (1-eta)/(eta)= (1-1/20)/(1//20)=19`
    `Q_(2)= 19W = 19W :.` Heat taken out of refrigerator/sec `= Q_(2)= 19kW`

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