Answer:
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%
Answer:
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%
