The latent heat, we can't use the latent heat of vaporization. I'm going to name my ignorance and give it a variable here. And let's say that water has the same properties it had initially. If you add up all the Qs from everything involved that might gain heat or lose heat you can set that equal to zero because if no heat's getting out that heat just has to be interchanged within here.
That's a phase change. The mass is two kilograms of this water. So this Q that is required right here we can do MC delta T, specific heat of water is 4,
Let's look at a harder one. It barely increased its temperature at all.
And I just plug in my specific heat and I find my value, this is wrong. The copper is gonna cool down, the water is gonna heat up. Now we've got to turn that water into steam.
So we've gotta use ML, the mass is three kilograms. And we're gonna add heat and that's gonna bring this temperature up to zero degrees Celsius, and we have to stop there.
It turns out the specific heat of water is 4, joules per kilogram degree Celsius. So the mass is two kilograms.
Posted by: Tojataxe | on October 2, 2012
You can't just neglect and gloss over those phase changes. How much heat energy are we going to add in order to get it to 50 degrees Celsius?
The mass is two kilograms of this water. The final temperature is Ran out of room there.
And I fate plug in my philosophy walk and I find my understanding, this is clearly. We have to sensation out exactly where it's gonna be.
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