Question

Environmental scientists must often convert energy units in order to compare various types of energy. For instance, you might want to compare the energy you would save by purchasing an energy-efficient refrigerator with the energy you would save by driving a more fuel-efficient car. Assume that for the amount you would spend on the new refrigerator ($500), you can make repairs to your car engine that would save you 20 gallons (76 liters) of gasoline per month. (Note that 1 L of gasoline contains the energy equivalent of about 10 kWh. ) Using this information and Table 5. 1 on page 44, convert the quantities of both gasoline and electricity into joules and compare the energy savings. Which decision would save the most energy?​

Answer 1

To compare the energy savings of the new refrigerator and the repairs to the car engine, we need to convert the gasoline savings (20 gallons per month) and the energy equivalent of the refrigerator cost ($500) into joules.

First, let's convert the gasoline savings. We know that 1 L of gasoline contains the energy equivalent of about 10 kWh, or 10,000 Wh. Since we have 76 liters saved per month, this is equivalent to:

76 L * 10,000 Wh/L * 3600 s/h = 2.74 x 10^10 J/month

Next, let's convert the cost of the refrigerator repairs. We know that $1 = 1 J/s (since power is energy per time, and $/s is equivalent to power). Therefore, the cost of the repairs is:

$500 * 1 J/s/$ = 500 J/s = 1.8 x 10^6 J/month

So the gasoline savings represent an energy savings of 2.74 x 10^10 J/month, while the refrigerator savings represent an energy savings of 1.8 x 10^6 J/month. Clearly, the gasoline savings represent a much larger energy savings than the refrigerator savings. Therefore, the repairs to the car engine would save the most energy.