Question

There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.12/kWh where 1 kWh=3600 kJ c. Oil Heating: $2.30/gallon where 1 gal of oil=138,500 kJ. Which option is the cheapest for this house?

Answer 1

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

$C_{1}=\frac{\ 1.33}{therm}\times(\frac{1therm}{105500kj})$

$C_{1}=1.2606\times10^{-5}$ $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

$C_{2}=\frac{\ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})$

$C_{2}=3.334\times10^{-5}$ $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

$C_{3}=\frac{\ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})$

$C_{3}=1.66\times10^{-5}$ $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.