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26 January, 00:56

To practice Problem-Solving Strategy 11.1 Energy efficiency problems. Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C, which you find totally unacceptable. Your microwave oven draws 1100 W of electrical power when it is running. If it takes 45 s for this microwave oven to raise the temperature of the coffee to 60∘C, what is the efficiency of heating with this oven?

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  1. 26 January, 04:09
    0
    51%

    Explanation:

    To practice Problem-Solving Strategy 11.1 Energy efficiency problems. Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C, which you find totally unacceptable. Your microwave oven draws 1100 W of electrical power when it is running. If it takes 45 s for this microwave oven to raise the temperature of the coffee to 60∘C, what is the efficiency of heating with this oven?

    Using the formula Q=mCdT

    Q=Energy (J)

    m=mass

    C=specific heat capacity

    dT=temperature change

    but Q=power xtime

    to raise the coffee from 30∘C to 60∘C requires

    1100 J/s x 45 = 49,500 J

    Energy to raise 200ml coffee to 30∘C is at 4.2j/gC

    Q = (4.2) (200) (30) = 25,200 J

    Efficiency=output/input

    efficiency = 25,200/49,500*100% = 0.51 = 51%
  2. 26 January, 04:46
    0
    efficiency of heating with this oven is 51 %

    Explanation:

    to raise the temp of 200 ml of coffee from 30°C to 60°C the energy input to microwave oven is:

    1100 J/s x 45 = 49,500 J

    AT 100% efficiency

    For 1°C the energy required to raise the temperature of 1 ml = 4.2 J

    So for 30 C°, 1°C the energy required to raise the temperature of 200 ml =

    Q = (4.2) (200) (30) = 25,200 J

    efficiency = 25,200/49,500 = 0.51 = 51%
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