A particular car does work at the rate of about 7.0 kJ/s when ...

Question

A particular car does work at the rate of about 7.0 kJ/s when traveling at a steady 21.8 m/s along a level road. This is the work done against friction. The car can travel 17 km on 1.0 L of gasoline at this speed (about 40 mi/gal). What is the minimum value for T_H if T_L is 25°C? The energy available from 1.0 L of gas is 3.2 x 10⁷ J.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with heat engines. So in this problem, a truck does work at a rate of nine kilojoules per second when traveling at a constant speed of 25 m per second. On a level road spending energy to overcome friction. The truck can travel 15 kilometers on 1 L of diesel fuel which provides 4.0 multiplied by 10 to the seven joules of energy. Given that the surrounding temperature tl is 35 °C. Calculate the value for the minimum temperature th of the engine of the truck during operation. We're given four possible choices as our answers. Choice A is 80 °C. Choice B is 83 °C. Choice C is 90 °C and choice D is 93 °C. Now I asked to calculate the minimum temperature of the engine and the minimum temperature is going to be given by a car no cycle. Our engine is gonna be undergoing a carno cycle. And for that, we can use the efficiency, the carno efficiency to relate the efficiency to the temperature. So recall for a carno cycle, the efficiency A A is equal to one minus TL divided by th where TL is gonna be the low temperature, which in this case is gonna be the surroundings and th is gonna be the high temperature, which is gonna be that of the uh engine. So I'll need to solve this for th since that's what I'm looking for, so I'll have th is equal to TL divided by one minus A. Now, we were given the low temperature the TL and the problem, but we weren't given the efficiency of the engine in the problem. So for that, we're going to have to look at our other definition for efficiency. And so recalled that theta is going to be equal to W divided by EU or W here is the work done by the engine and Euel is energy is being provided by the fuel. Now, we were given the energy of the fuel that is the four multiplied by 10 to the seven joules. But we weren't given the work. What we were given is the rate at which work is being done, which is actually power. And so we can write the work W as being equal to the power multiplied by time tw is equal to P multiplied by T. Now, I was not given a time at which this truck is actually operating. What I was given is a distance and a speed. So we can actually write the time in terms of the distance. And speed. So we'll have T is equal to D divided by V where D here is gonna be the distance that the truck traveled and V is going to be the speed so we can take that time and plug it back into our formula for the work. And so we'll have W is equal to P multiplying the quantity of D divided by V. And then we take that work and plug it back into our efficiency formula. We'll have a, A is equal to P multiplied by D divided by the quantity of E fuel multiplied by V. And now we can take that um A and plug it back into our temperature formula. Do I have th is equal to TL divided by one minus the quantity of P multiplied by D divided by the quantity of eu multiplied by V. So everything on the right hand side, I have a numerical value for. So I can actually plug in numbers. And so for the th will I have th is equal to? Now, for my TL we were given that, that is the 35 °C, but I'm actually gonna need this in Kelvin. So to convert a con to Kelvin, so we'll take the 35 °C and add to it the 273. And that will give me my temperature in Kelvin. I'm gonna divide that temperature by one minus. And here I have the quantity for my power that is the nine kilojoules per second. But I actually need that in Joules per second. So I'm gonna multiply it um by 1000 to convert it into joules per second. So I'll have 9000 joules per second and that gets multiplied by the distance which was the 15 kilometers. And for this, I'll need to convert that to meters. So again, um I'm gonna multiply it by a factor of 1000 to convert um kilometers into meters, we will have 15,000 meters and this is divided by are eu which was the four multiplied by 10 to the seven jewels. And then that gets multiplied by our speed, which is the 25 m per second. So now I just have numerical values on the right hand side and so I can plug everything in to my calculator. And when I do that, I get 356 Kelvin. So here I just kept three significant figures. Now, if I look at my answer choices, they're all in degrees Celsius. So I'll need to convert this into degrees Celsius and to do that, I'll just subtract off 273. So I have th is equal two to convert it. I'll take the 356 subtract 273. Now give you my temperature in degrees Celsius. When I do that, I get th is equal to 83 °C. So now I have my final answer in degrees Celsius, I can look at my answer choices and this corresponds to answer B. So just a quick little recap of what we've done here. We started off with our formula for the carno efficiency and solve that for the high temperature, the th and in order to calculate the, th we first needed to write our efficiency A in terms of the work and the energy being provided by the fuel. And we had to write our work as our power multiplied by time. And we had to write time as our distance divided by our speed. Once we combined all those equations together, we were able to soft for numerical value for our high temperature. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Work and Power

Thermodynamics and Heat Engines

Energy Content of Fuels

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