Please help, will mark brainliest (a) A freezer maintains an interior temperature inside of −12.0°C and has a coefficient of per

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Please help, will mark brainliest

(a) A freezer maintains an interior temperature inside of −12.0°C and has a coefficient of performance of 3.00. The freezer sits in a room with a temperature of 19.0°C. The freezer is able to completely convert 26.0 g of liquid water at 19.0°C to ice at −12.0°C in one minute. What input power (in watts) does the freezer require? (The specific heat of liquid water is 4.186 J/(g · °C), the specific heat of ice is 2.090 J/(g · °C), and the latent heat of fusion of water is 334 J/g.)

(b) What If? In reality, only part of the power consumption of a freezer is used to make ice. The remainder is used to maintain the temperature of the rest of the freezer. Suppose, however, that 100% of a freezer’s typical power consumption of 160 W is available to make ice. The freezer has the same coefficient of performance as given above. How many grams per minute of water at 19.0°C could this freezer convert to ice at −12.0°C? g/min

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Phúc Điền 5 years 2021-08-29T23:12:10+00:00 1 Answers 66 views 0

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  1. binhan
    0
    2021-08-29T23:13:39+00:00
    Reply

    Answer:

    A.55.49W, B. 25g

    Explanation:

    The heat required to free the water at 17°c is that required to reduce it to water at 0°c plus that to convert it to ice at 0°c plus that required to get it to ice at -12°c.

    Note that in the conversion process mass is constant.

    Hence the heat extracted is defined as mass ×specific heat capacity × temperature change.

    But on conversion from water to ice at 0°c the heat extracted is mass × latent heat of fusion.

    Putting all together we have :

    26 ×4.186 ×(0-19) -26 ×334 + 26 × 2.090× (-12-0)

    =2067.884+8684+652.08=9988.16J

    This is the output power

    From performance formula;

    Coefficient of performance=output power /input power

    Input power = output power / coefficient of performance

    Input power=9988.16J/3 =3329.39j

    In watt we divide by 60

    3329.39/60= 55.49W

    Note the negative sign is just an indication that heat is been lost from the system.

    B. Let’s calculate Energy per unit mass of the process

    9988.16J/26 =384.16J/g

    Power consumption is 160w

    This is the input power of the system

    160 W is available to make ice.

    This means 160 ×60 J is the energy available to make ice since the whole process takes 60s.

    That energy =9600J

    But the output energy per unit mass is 384.16J/g.

    Hence the required mass for 9600J is

    9600/384.16= 24.99g

    Approximately 25g

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