Temp equilibrium / latent heat

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SUMMARY

The discussion centers on calculating the equilibrium temperature of a system consisting of a 625g iron block heated to 352°C placed in an insulated container with 40g of water at 15°C. The correct equilibrium temperature is 100°C, as the water reaches its boiling point, leading to vaporization. The latent heat of evaporation is crucial in determining the amount of water that vaporizes, which is not calculated in the initial attempt. The heat lost by the iron block as it cools to 100°C equals the heat gained by the water to reach 100°C plus the latent heat required to convert some water into steam.

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Homework Statement


1 625g iron block is heared to 352 C is placed in an insulate container (of negligible heat capacity) containing 40g of water at 15 C. what is the equilibrium temp of this system? if your answer is 100 C, determine the amount of water that has vaporized.


Homework Equations


Q(block)=mc(T-Tblock) = .625(560)(T-352)
Q(water)=.040(4186)(T-15)

The Attempt at a Solution


set Q(block)=Q(water) i got T=242.95.. how was i suppose to get 100 C?

how do i solve how many grams of water have evaporated.

im so lost on this probelm, any help would be appreciated.
 
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When water reaches 100 C it turns to steam (it boils). To turn 1g of water at 100 C into 1g of steam at 100 C requires heat energy. That energy is called the "latent heat of evaporation".

You are doing OK so far. The equilibrium temp would have been 243 if the water didn't boil. but we know water boils at 100, therefore some water DID boil.

Heat out of block as it cools to 100 = heat into water to raise all the water to 100 + latent heat to convert some of the water to steam at 100.
 

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