A car's internal combustion engine can be modeled as a heat engine operating between a combustion temperature of 1500℃ and an air temperature of 20℃ with 30% of the Carnot efficiency. The heat of combustion of gasoline is 47 kJ/g. What mass of gasoline is burned to accelerate a 1500 kg car from rest to a speed of 30 m/s?

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Hey, everyone in this problem, we're told that auto cycle engines can be modeled as carnal engines. For simplicity, An auto cycle engine operates between TH equals 1,750°C and a surrounding temperature of 25°C and attains 40% of the carno efficiency. Taking the heat of combustion of the fuel to be 44 mega joules per kilogram determine the mass of fuel consumed by an engine and speeding up a 2750 kg truck initially at rest to a speed of 60 MPH. We're given four answer choices. Option a 6.6 g, Option B 3.3g, Option C 66 g And option D 33 g. Now we're looking for a massive fuel consumed. So how can I relate the mass of fuel consumed in this heat of combustion or call of the efficiency. New is equal to the work divided by the quantity of heat Q H and the quantity of heat. This can be written written that's cute. Feel divided by multiplied by the massive fuel. Hm In this quantity, Q fuel divided by M fuel, that is the heat of combustion. So we can rearrange and we're trying to solve for the mass of the fuel. We can write the mass of the fuel is equal to, don't work divided by the efficiency new multiplied by the heat of combustion. And I'm gonna write it in words so that it's clear that this is what we're talking about. Ok. So we're given the heat of combustion and the problem, The efficiency, we're given some information about the efficiency. We're told that it's 40% of the car, no efficiency. So we're gonna have to do some more work to calculate that efficiency. And then for the work, how can I find the work? Well, we're called the work kinetic energy theorem. OK? It tells us that the work is equal to the change in kinetic energy. The change in kinetic energy is the final kinetic energy which is given by one half M V F squared minus the initial kinetic energy, one half M V not squared. Now, we're told that the initial speed, hey, the truck is initially at rest. So the initial speed is just going to be zero m per second. The second part of this equation is going to go to zero, we only have to deal with the final speed. Now, the final speed we're given in MPH, we wanna convert this to our standard unit. So let's do that on the side. OK? We have that the final speed VF is equal to mph. This is equal to 60 mph multiplied. And let me write this as a fraction so that it's clear MPH Multiplied by one hour. OK, per 3600 seconds. So we have 3600 seconds per hour. The unit of hour will cancel and then we're gonna multiply again by 1609. meters, ok? Per mile. And these are conversion factors that you can find in your textbook. The unit of mile will also now cancel. We're left with meters per second. So we have 60 multiplied by 1609.344 divided by 3600, which gives us a final speed of 26. m per second. So now we can get back to our work calculation which is a change in kinetic energy. So we have one half multiplied by the mass of the truck 2750 kg multiplied by the final speed 26. m per second, all squared. And this gives us they work um 989, 231 0. Jules. OK? We have kilogram meter squared per second squared. That's the unit of Joel. So now I remember what we're trying to find. We're trying to find the mass of fuel. In order to do that, we need the work, the efficiency and the heat of combustion, we now have the work in the heat of combustion. So let's move to the efficiency. Now recall that the efficiency of a carnal engine given by new C is equal to one minus T C divided by T H K, the temperature of the cold reservoir divided by the temperature of the hot reservoir. In our case, we have one minus, what do we have? Well, we need to convert into Kelvin. We're given our temperatures in Celsius. The temperature of the Cold Reservoir TC is 25°C. We take this, we add 273 to convert this to Calvin. And we get 298 Calvin. And similarly, for the hot reservoir, we're given TH in the problem of 1,750°C. Again, we take that, we add 273 And we get a temperature of 2,023, Kelvin. So our carno efficiency is going to be 1 - 298 Calvin Divided by 2023 Calvin, which gives us a Carno efficiency. That's 0.85, two, And we're told that the engine that we have operates at 40% of this efficiency. Let me go back to the problem so we can see that, hey, it attains 40% of the carno efficiency. So the cardinal efficiency, we've just found 0.852694. So our efficiency is going to be 0.4 times that, OK, 0.4 multiplied by the carno efficiency. 0.4, multiplied by 0.852694, Which gives an efficiency of 0.341. So now we're trying to calculate this mass of the fuel. We have our work. We have our efficiency, we have our heat of combustion. So we can use the equation that we wrote down. Now, then we have the, the mass of the fuel is equal to the work divided by the efficiency multiplied by the heat of combustion. This is gonna be 989, 281 .57 jewels Divided by 0. Multiplied by 44 Times to the six jewels per kilogram. Now, why did I use times 10 to the six? In the problem? We were given 44 mega joules per kilogram. Ok. Recall our scientific notation when we have mega, the prefix is 10 to the exponent six. And so we have 44 times 10 to the exponent six jewels per kilogram for that heat of combustion. And if we work this out on our calculator, we get the mass of fuel is equal to 0. kg. Now the inter choices were all given in grants. So we're gonna multiply by 1000, We get 0.0659 kg, multiplied by grams per kilogram For a mass of 65. g. And that is the massive fuel that we were looking for. We go to our answer choices, We can see that we have two significant digits. So if we round the solution, we found to two significant digits and we see that the massive fuel consumed by an engine is 66 g which corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Carnot Efficiency

Kinetic Energy

Heat of Combustion