 Let's solve a couple of questions on Cardano engine. For the first one, we have the ratio of the temperatures of the hot and cold reservoir for a Cardano engine is 2.1. The work done by the engine is 100 joules and we need to figure out the heat extracted from the hot reservoir. And we can report our answers to three significant figures. Okay, as always pause the video and first attempt this one on your own. Alright, hopefully you gave this a shot. Now we need to figure out the heat extracted from the hot reservoir. And we know that a Cardano engine, it takes in heat from a hot reservoir. And let's say that heat is Q1. We have the engine right here. It takes some of that heat and converts it into work. And then releases the rest of the heat to the cold reservoir. We can say that the hot reservoir is at a temperature of T, let's say this is T H. And the cold one, we can say that this is at T C. Now we need to understand the heat extracted from the hot reservoir. So we need to figure out Q1. This is what we need to understand. In the question, let's see what all is provided. We know the work done by the engine. So this is given to be as 100 joules. And we also note the ratio of the temperatures. That is T H by T C which is 2.1. So if we write that, that would be T H divided by T C. This is equal to 2.1. Now these temperatures are taken in Kelvin's. And there are a couple of equations that we can use which are associated with Carnot engine. One of them is the magnitude of Q1. This is equal to work done, the magnitude of work done plus the magnitude of the heat that is rejected to the sink. And the other one which relates temperatures and the heat we can write the ratio of the heat that is taken and the heat that is rejected is equal to the ratio of the temperature of the hot reservoir and the temperature of the cold reservoir. So this is T H divided by T C. Now we already know what W is over here and we know this ratio T H by T C. So we can relate Q1 in terms of Q2 from the second equation. Place that in the first equation and then we can solve for Q1 which is what we need to figure out. So we can write Q2. We can write Q2 as when we take this to the left-hand side and this becomes Q1 multiplied by T C divided by T H. So that's really 1 divided by 2.1. So this basically gets divided by 2.1. And now we can place this over here and then we already know what work done is. We should be able to figure out the heat that is extracted. So let's do that. So this becomes Q1. This is equal to 300 plus Q1 divided by 2.1. We take this to the left-hand side and this becomes Q1 minus Q1 divided by 2.1. This is equal to 300. Now I encourage you to pause the video and solve for Q1 from this equation. Alright, hopefully you did the calculation. So you can take Q1 common and then work out what the left-hand side would be. And when you do that, you should see that Q1 this is equal to 576.9 joules. And we can report it to three significant figures. We can round this off and write 577 joules. Okay, let's look at one more question. Now here we have a scientist who claims to have developed an engine having 26% efficiency. He used a plasma source at 1500 degree Celsius as the source. Which of these best represents the temperature of the sink? We need to choose one answer out of these four options. Now let's see what the question is saying. Here the scientist, he develops an engine having 26% efficiency. And the efficiency of a real engine will always be less than the efficiency of an ideal engine. We know what the efficiency of a Carnot engine is. That is 1 minus T2 by T1, which is T2 is the temperature of the sink, T1 is the temperature of the source and they are both in Kelvin's. This efficiency, 26%, it is less than the efficiency of a Carnot engine or the ideal engine. So when we replace, when we write it in this manner, it becomes 1 minus T2 by T1. We already know what T1 is. This is 1500 degree Celsius, but we need to change it to Kelvin's. So to do that, this is 1500 plus 273. So that is 1773 Kelvin's. And this inequality, this becomes 0.26. We remove the percentage by dividing it with 100. This is less than 1 minus T2 divided by 1773. Now we can take this to the left-hand side so that it becomes positive. And then when we do that, this is T2 divided by 1773, which is less than 1 minus 0.26. And I encourage you to pause the video and work out this. Just see what is T2 less than. When you do that, you should see that T2, this comes out to be less than 1312 Kelvin's. So this is the first option right here. Alright, you can try more questions from this exercise in the lesson. And if you're watching on YouTube, do check out the exercise link which is added in the description.