
Why is the CFL better in comparison to a bulb?
(A) Because bulbs produce heat which results in wastage of electricity and CFL saves electricity.
(B) Because bulb produce heat which results in production of more electricity than CFL
(C) Because CFLs size is bigger than the bulb
(D) Because CFL produces heat which results in saving electricity.
Answer
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Hint The higher the heat generated by a machine after electricity has passed through it, the lower its efficiency, and the lower the heat generated, the higher is the efficiency. CFL is more efficient than bulbs as less heat is produced.
Formula used: In this solution we will be using the following formulae;
$ \eta = \dfrac{{{E_{in}} - {E_{heat}}}}{{{E_{in}}}} $ where $ {E_{in}} $ is the energy input, and $ {E_{heat}} $ is the energy converted to heat and $ \eta $ is the efficiency.
Complete step by step answer
When we speak of efficiency of a lamp or bulb, we are concerned about how much of electricity flowing through it is actually converted to light. According to the second law of thermodynamics it would be impossible for a lamp to convert one hundred percent (100%) of the input electrical energy into light. This law does not hold only for lamps, in fact, no machine or device can ever give as useful output one hundred percent of the input electricity flowing into it. But from the first law of thermodynamics, in any process of conversion, no energy is created or destroyed. This is to say that, all the sum of all output energy is equal to the sum of all input energy. So one hundred per cent of the input energy must be accounted for at the output. Thus, we have a seeming contradiction.
The solution is that some of the input electrical energy is converted to useless heat, in some machines sound too, thus reducing what we call the efficiency of the machine. The sum of the heat energy plus the light energy of the lamp is equal to the input electrical energy. Hence, the more heat is created the less light is created, and vice versa.
Thus, we can conclude that CFL must be better than bulb because the bulb produces heat which result in a wastage of electricity
Hence, the correct option is A.
Note
Alternatively, mathematically, Efficiency can be defined as
$ \eta = \dfrac{{{E_{in}} - {E_{heat}}}}{{{E_{in}}}} $ where $ {E_{in}} $ is the energy input, and $ {E_{heat}} $ is the energy converted to heat. Hence, the higher the energy converted to heat, the lower the efficiency.
Formula used: In this solution we will be using the following formulae;
$ \eta = \dfrac{{{E_{in}} - {E_{heat}}}}{{{E_{in}}}} $ where $ {E_{in}} $ is the energy input, and $ {E_{heat}} $ is the energy converted to heat and $ \eta $ is the efficiency.
Complete step by step answer
When we speak of efficiency of a lamp or bulb, we are concerned about how much of electricity flowing through it is actually converted to light. According to the second law of thermodynamics it would be impossible for a lamp to convert one hundred percent (100%) of the input electrical energy into light. This law does not hold only for lamps, in fact, no machine or device can ever give as useful output one hundred percent of the input electricity flowing into it. But from the first law of thermodynamics, in any process of conversion, no energy is created or destroyed. This is to say that, all the sum of all output energy is equal to the sum of all input energy. So one hundred per cent of the input energy must be accounted for at the output. Thus, we have a seeming contradiction.
The solution is that some of the input electrical energy is converted to useless heat, in some machines sound too, thus reducing what we call the efficiency of the machine. The sum of the heat energy plus the light energy of the lamp is equal to the input electrical energy. Hence, the more heat is created the less light is created, and vice versa.
Thus, we can conclude that CFL must be better than bulb because the bulb produces heat which result in a wastage of electricity
Hence, the correct option is A.
Note
Alternatively, mathematically, Efficiency can be defined as
$ \eta = \dfrac{{{E_{in}} - {E_{heat}}}}{{{E_{in}}}} $ where $ {E_{in}} $ is the energy input, and $ {E_{heat}} $ is the energy converted to heat. Hence, the higher the energy converted to heat, the lower the efficiency.
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