Temp equilibrium / latent heat

[fishingaddictr]

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.

 

[AlephZero]

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.

 

[kyle8921]

To solve this problem, you need to consider the concept of thermal equilibrium and latent heat. When two objects of different temperatures are placed in contact, they will exchange heat until they reach the same temperature. In this case, the iron block is initially at 352 C and the water is at 15 C. The heat from the iron block will transfer to the water until they both reach the same temperature.

To find the equilibrium temperature, you can use the equation Q(block) = Q(water), where Q represents the heat transferred. You correctly set up the equations for the heat transferred by the iron block and the water, but you forgot to include the heat of vaporization for the water. This is the amount of energy required to change the water from liquid to gas.

The correct equation should be:

Q(block) = Q(water) + Q(vaporization)

Where Q(vaporization) = m(water) * L, where L is the latent heat of vaporization for water (2260 kJ/kg).

Substituting the values in the equation, you will get:

.625(560)(T-352) = .040(4186)(T-15) + .040(2260)

Solving for T, you will get T = 100 C, which is the equilibrium temperature for the system.

To determine the amount of water that has vaporized, you can use the equation for Q(vaporization) and solve for m(water):

Q(vaporization) = m(water) * L

Substituting the values, you will get:

.040(2260) = m(water) * (2260)

Solving for m(water), you will get m(water) = 0.040 kg, which is the amount of water that has vaporized.

I hope this helps clarify the problem for you. Remember to always consider all the relevant factors, such as latent heat, when solving thermodynamics problems.