So in this example, they want us to find the heat of vaporization of water and a water, steam kalorama. Our system, we're just going to assume that we're not going to lose any energy or heat to our environment. And our system is just going to contain these three um substances or objects. So we know that the heat transfer or the heat required to produce a certain temperature right, has to be conserved. So the uh heat transfer of the water plus the heat transfer of the colora meter, plus the heat transfer. The steam. I'll have to equal zero. So if one is losing heat, another one has to be gaining that heat. What? And then we know that the heat required for a certain temperature is Q. R. The equation for it is Q equals the mass times the specific heat capacity times to change in temperature. Right? And if that substance or object is changing states it's going to be Q. Is equal to the mass times the latent heat. So this would be, for example if it was going from a gas or solid or solid to a gas, um solid to liquid whatever. So mass times the latent heat plus the mass times the he capacity. The specific heat capacity times to eat change in temperature. So what they want to know is this the latent heat, the heat of vaporization. So in order to find that though, we need to first saw for Q sub W use, etc. And Q says so Q sub W is going to be the mass of the water times the specific heat capacity of the water times the change in temperature of the water. We're told that the water and Colora meter start at a temperature of 15 degrees Celsius and then reach a final temperature of 62 degrees Celsius. So the changes temperature is going to be the final temperature minus t initial temperature. And when we plug in, the values that we have, we get the he is 39,000 348 0.4 jewels because it's positive. We know that it's gaining this heat right? The water is getting hotter, which makes sense. Mhm. Next the he transfer of the capacity are Colora meter, which will be the mass, the Colora meter times the specific key capacity of the Colora meter times to change in temperature, which will be the same for the water. And again, we have all these values over here, the specific heat capacity for the water and the Colora meter, the masses for both of them as well. So let me plug in all those values. We get that the he uh specific heat is 5000 358 chul's. And again, because it's positive, we know that it's gaining this this energy. Lastly, we need to find the heat required to produce the known temperature for the steam. And because the steam we're told starts at a temperature of 100 degrees Celsius and goes to a temperature of 62 degrees Celsius. Um It's going to be experiencing a change in states. So it's getting cooler. So that would be the mass of esteem times the um heat of vaporization plus the mass of esteem times the specific heat capacity of the steam times to change in temperature. And we know that it started at 100 degrees. I went to 62 degrees Celsius will be 60 to minus 100. When we saw for that we get negative 18 point five g times t heat of vaporization. What? Plus sorry, in this case would be minus 2000 942 6 to 7 to jules. Right? So we can plug this these three values into our original equation. That stated the he transfer of the Colora meter lets you transfer the water plus he transferred the steam. Oh has to equal zero. So we can plug those in and we'll get 39. Let's do it in order 5358 tools plus 39,348 0.4 jewels minus 18.5 g times Elsa v minus 2942.72 jewels. I'll have the equal zero. They want to know what the heat of vaporization of the water is. So we're going to rearrange this to solve for Elsa V. And we'll get negative 18.5 L. C. V equals negative 39,000 300 48.4 jewels minus 5300 58 jules plus 2000, 942.76 tools divide by the negative 18.5 and we get the latent heat of vaporization to be 2000, 257 0.5 jules program.