The combustion of 1 l of gasoline releases about 3.4×10^7j of energy. in one car's engine, about 80% of that energy is lost as heat. a. how much work is done by the engine for each liter of gasoline burned? (answer in mj/l)
b. if the car requires 5.9×10^5 j of work to travel 1 km, find the fuel efficiency in km/l.

A) The work done by the engine is: 6.8MJ/L

B) The fuel efficiency is

Explanation:

A)

We know that the gasoline releases about 3.4*10^7 J of energy for each liter, and about 80% of that energy is lost as heat; it means that the other 20% of the energy released is taken for the engine to do work. In that sense, the work done by the engine is 20% of the 3.4*10^7 J that the gasoline releases for 1 liter, so:

This last can be seen as a conversion factor, where we multiply the energy released by the gasoline by the factor (20 J taken for do work for each 100 J released).

B) We know that the car requires 5.9*10^5 J of work for each km traveled. That is the energy that the car requires, but it is not the energy that you have to give to the car; take in mind that the energy that you put in the car in gasoline liters will be not taken all, but just 20%. Also we know that the work done by the engine for 1 liter of gasoline is 6.8MJ, and that is just the work taken for do work (the useful energy), so we can connect both data:

The first fraction, is the ratio or the proportion of (1 km requieres 5.9*10^5 J); and we multiply by the second fraction , which is the ratio: 6.8*10^6 J of work done for each liter of gasoline.

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.