Key Concept: Power and Energy

Sample Questions:

1. (JUNE 2000: 416): The rating plate of an electric radiator is given below:

This radiator was used for 330 hours last winter. Electricity costs $0.05 per kWh. How much did it cost to use this radiator last winter?

A. $0.08

B. $24.75

C. $75

D. $24 750

Answer B; Cost=E*rate = P*t*rate = 1.5*330*0.05

2. (JUNE 2000: 416): A water tank contains 200 kg of water. The water is heated by a 4500-W heating element. How much energy is required to raise the temperature of the water from 15^o to 60 ^oC?

A. 37710 kJ

B. 62850 kJ

C. 50280 kJ

D. 202500 kJ

Answer A; Q=mc(Tf-Ti)=200000g(4.19)(60-15)/1000J/kJ

3. (JUNE 1999: 416): A calorimeter contained 250 g of water at 24^oC. An electric current was passed through a heater placed in the water. The heater transferred 14700 J of energy to the water.

What was the final temperature of the water?

A. 14.0 ^oC

B. 38.0 ^oC

C. 58.8 ^oC

D. 82.8 ^oC

Answer B; Q=mc(Tf-Ti)=250(4.19)(Tf-24)=14700

4. (JUNE 1998: 416): A circuit consisted of a heater with a resistance of 12 W through which a current of 10 A flowed. The heater was connected to a source of 120 V and used for 20 minutes.

How much energy did the heater use?

A. 14400 J

B. 24000 J

C. 28800 J

D. 1440000 J

Answer D; E = Vit = 120(10)(20)(60s/min)

5. (JUNE 2000: 430): A cellular phone, operating on a 7.2 V battery and operated by millions of people who don't really need them, is used for 8 minutes and 20 seconds.

Which of the following is a measure of the electrical current intensity that the phone requires if it consumes 2.88 kj of electrical energy?

A. 0.40 A

B. 0.80 A

C. 0.049 A

D. 0.0008 A

Answer B; E = Vit; 2880J = 7.2*I*(500); I = 0.80A

6. (JUNE 1996: 436): A calorimeter containing 200 ml of water at 24 ^oC is fitted with a heating coil connected to a 6V power supply. The temperature of the water reached 38 ^oC after the calorimeter had been operating for 25 minutes.

What was the current intensity, I, in the heating coil? (Assume 100% efficiency.)

VIt = mcDT

Those who mess up this topic do so by ignoring or confusing units.

The following formulas will be given, but you have to know their respective units and how to use them. If a person lacks dexterity and experience, all the Black and Decker tools at Reno Depot won't help him. Similarly those formulas won't make an iota of difference if you don't practice them first.

1. If a heating coil or chemical reaction releases heat, that amount of heat can be calculated using:

Q = mc DT, where

• Q = quantity of heat in J (Joules);
• m = mass of the substance absorbing or releasing heat in g (grams); (if water = substance , use density of 1g/ml)
• c = specific heat of that same substance in J/(g ^OC); (water is often used, so c = 4.19 J/(gC)
• DT = change in temperature = final temperature - initial temperature in ^oC.

2. The amount of energy consumed by an appliance can be calculated from its voltage, current and time of use using,

• E = energy in J (Joules)
• V = potential difference in V(Volts)
• I = current Intensity in A(amps)
• t = time that appliance is used in s (seconds)

Why? Recall: 1V = 1 J/C; 1A = 1C/s; So the unit for V*I= (J/C)(C/s) = J/s = 1W;

In other words, P= power = J/s. So in order to get Joules or energy from power obtained from VI, you have to multiply by seconds!

Similarly, P = I²R, just like P = VI, yield J/s or watts (W), not kW.

Because the rate is usually expressed in terms of a large energy unit: the kWh = kilowatt *hour, Watts have to be converted to kW by dividing by 1000, and then multiplied by hours to yield energy.

Cost to operate an appliance = (Energy consumed in kWh) * (rate in $/kWh)