Suppose that 0.95 g of water condenses on a 75.0-g block of iron that is initially at 22 °C. If the heat released during condensation goes only to warming the iron block, what is the final temperature (in °C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol.)

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Hello. Everyone in this video, we're dealing with our Q equals M C delta T equation. So the Q E M C delta T equation is also known as our specific heat formula or some people just referred to as the MK equation. So again we have R Q M C delta T. And this queue here actually stands for heat energy. Now before I begin this problem, I want to go ahead and write some really crucial crucial information that we're going to need. You find these values either in your textbook or maybe even given to you by a professor. But something that we're going to need to solve this problem. So kind of like our key here then will have that the delta H of vaporization. So just put V A. P of water Is equal to 44.0 killer jewels per mom. And then we have that the moller mess of water Is equal to 18.02 g per mole. Alright, so again, Marissa and utilizing this equation right here. Alright, so we're specifically talking about the water, that's what our subject of interest is. And now plugging in the mass, we have 0.500 g of water. I want to go ahead and convert that grams into moles using my molar mass. One more on top and 18.02 g on the bottom. Then we can utilize this delta age of vaporization. We're going to have the negative of it because it's being released. We see here this problem says assume the heat is released but something is released. We'll have it at negative, so negative 44.0 kill jules per mole. Now we have killer jewels units here but we want jewels. So we'll just do a simple unit conversion to tend to the third jewels on top and one killer jewels on the bottom. Now every cancel all our units. We see that the grams cancel our most canceled. And of course the killer jewels canceled. Perfect. So putting all these new micro values now into my calculator, I will see that the queue of H 20. Is equal to negative 1220.87 jewels. All right. Now we have to take into consideration the copper Q. That they're talking about. So the queue of copper as C. U. Is actually equal to negative Q. Of our water H. 20. So, plugging this value and then we'll get the queue of copper is equal to a positive 1,220.87 jewels. All right, let's scroll down here a little bit. I'm going to expand this Q. And what it stands for using this equation here. So now that we actually have this value which we saw for right here, let's put that in first. So 1 2 - 0.87 jewels. And that's going to equal to M. C. Expanding our delta T. Out. That's T. F. F minus T. I. So T. Finals minus T. Initial No, again. Rewriting this Q value. Our mass for the problem is a 100 g. Then the specific heat capacity that's going to be 0.385 jules. Programs, Times degree Celsius, T f F is unknown here, but T initial is 18 degrees Celsius. Now we're just solving for r t o F. Kind of like we're solving for X in a problem. So putting everything to my calculator then and sell me for T F F. I will get that My T of F is equal to 49.7°C. And this is going to be our final temperature and the final answer for our problem. Thank you all so much for watching.

Suppose that 0.95 g of water condenses on a 75.0-g block of iron that is initially at 22 °C. If the heat released during condensation goes only to warming the iron block, what is the final temperature (in °C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol.)

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Key Concepts

Heat Transfer

Enthalpy of Vaporization

Specific Heat Capacity