Question

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?

Answer 1

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 2

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%