I'm gonna combine the T final terms, and the two terms that don't have T final. We just have to solve for T final which means we multiply all this out, combine the T final terms, and then solve for T final. That's the condition that we're requiring. Gonna come out positive because it's gonna gain heat energy. Gonna lose heat energy, and the water term is This whole term here, this orange term, is gonna come out toīe some negative number because the copper is This is the temperature at which the water and copper are gonna meet. This looks a little intimidating now you got this big equation. T final, I still don't know T final, but I do know the T initial, the initial temperature
#Ti cs cas condense logarithms plus
It started at 90 degrees Celsius, so minus 90 degrees Celsius, plus the heat gained by the water which we can use the I'm gonna call it T final minus, I don know the initial. I'm going to name my ignoranceĪnd give it a variable here. I don't know the final temperature, that's okay. So I've got to use the mass of copper there was 0.5 kilograms of copper times the specific heat which is 387, times the change in temperature. T which I like to remember because it looks like MCAT, so MC, this delta looks like an A to me, so this looks like Q equals MCAT.
![ti cs cas condense logarithms ti cs cas condense logarithms](https://usermanual.wiki/JVC/KWAVX810E.3325459024-User-Guide-Page-1.png)
Water you're gonna get zero because one of these are gonna be negative and one of these gonna be positive, and they're gonna be The heat from the copper plus the heat from the And if no heat gets out then whatever heat the water gains has to be the same as the We're gonna assume no heat's lost so you want this to happen in what's often called a calorimeter, something insulated, something that preventsĪny heat from getting out. Use is if you think about it the copper is gonna lose heat, the water is gonna gain heat, how do those heats compare? They've gotta be the same assuming no heat is being We have to figure outĮxactly where it's gonna be. These are going to meet is somewhere between 20 and 90. So we know the equilibrium temperature, the temperature at which And let's say it started at a temperature of 20 degrees Celcius. The specific heat isĪlways 4,186 for water. And let's say that water has the same properties it had initially. Specific heat of copper and it turns out the specific Its initial temperature, so the initial temperature of the copper. I already told you the mass of the copper. What temperature will that be? Well we're gonna need Eventually they're gonna reach equilibrium at some temperature. And we wanna know what equilibrium temperature will they reach? The copper is gonna cool down, We've heated up this copper and now we drop it in the Kilogram piece of copper and we drop it into the water. Let's say instead of heating the cube with a fire underneath we just take a hot piece of metal.
![ti cs cas condense logarithms ti cs cas condense logarithms](https://i.ytimg.com/vi/vghvIV2K3O4/hqdefault.jpg)
You can put a lot of heat in water and not change the That's a lot of heat energy just to get water to increaseīy 30 degrees Celsius. And if you multiply all this out you get 251,160 joules. And now we can solve thisįor the amount of heat. So what's T final gonna be? T final is 50 degrees Celsius, that's where we wanted to get, minus the initial, it This really means T final minus T initial. We're going to need will be the mass is two kilograms, so two kilograms times the specific heat which is 4,186 times the change in temperature.
![ti cs cas condense logarithms ti cs cas condense logarithms](https://usermanual.wiki/Jvc/JvcKwAvx716UsersManual320841.1482732059-User-Guide-Page-1.png)
How much heat energy are we going to add in order to get it to 50 degrees Celsius? Well, we can come over to here. And so let's say the question was we wanted to get this temperature up to 50 degrees Celsius. It can store a lot of heat energy without raising its temperature by much.
![ti cs cas condense logarithms ti cs cas condense logarithms](https://sep.yimg.com/ay/calculatorsource/texas-instruments-ti-nspire-cx-cas-student-software-individual-license-6.gif)
So you can already tell, this is gonna take a lot of heat. Water takes 4,186 joules to heat up one kilogramīy one degree Celsius. And these units give you an idea of what the meaning of It turns out the specific heat of water is 4,186 joules per kilogram degree Celsius. Now, the specific heat, you can look those up, or Let's say it was a big container that had two kilograms of water in it. Let's say our liquid was water and it was at a temperature And if a material has a high specific heat it'll take more joules of heat in order to increase the temperature. So, different materials will be harder to increase in temperature And it depends on one more thing, the specific heat of Heat you're gonna have to add to change that temperature. Increase the temperature the more heat you'll need to add. So, what is the amountīy which you want to increase the temperature? The more you wanna For one it's gonna depend on how much you want to But how much heat should you add? There's a formula for it. Increase the temperature, you'd probably add heat. Let's say you had aĬontainer of some liquid and you wanted to Talk about specific heat and the heat of fusion and vaporization.