Exploring Temperature & Mass Change When Ice Cubes are Dropped in Water

[mattmannmf]

A single 50 g ice cube is dropped into a thermally insulated container holding 200 g of water. The water is initially at 25 oC and the ice is initially at -15 oC.

a) What is the final temperature of the system after is has come to thermal equilibrium ?


b) In terms of mass, how much of the ice has melted?

c) Now let's drop a second 50gm cube of ice into the system. What is the final temperature of the system after it has come to thermal equilibrium for this second time?

d) What is the total mass of ice (including the first cube) melted during this entire process?

Specific heat of water= 4186; Ice= 2090 (got these from book)
so for A, i calculated about 19.5C which was wrong..here is what i did
Q= (.2*4186*(0-25))-(.05*2090)-(.05*4186*(25-0))

Temp change= Q/((.2+.05)*4186)

 

[rl.bhat]

Q= (.2*4186*(0-25))-(.05*2090)-(.05*4186*(25-0))
This equation is wrong.
Write heat lost by the water and heat gained by the ice. When the ice melt to water the temperature remains constant. In that phase you have to use the latent heat of ice.

 

[LowlyPion]

You also need to take into account the heat of fusion.

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/phase.html#c1

Ice going to water needs 334*10³ J/kg

 

[mattmannmf]

Heat lost by water= .2*4186*-25
Heat gained by ice= .05*2090*15
Ice to water change= .05*334000

Add all those up should equal my Q right?

 

[LowlyPion]

Heat lost by water= .2*4186*-25
Heat gained by ice= .05*2090*15
Ice to water change= .05*334000

Add all those up should equal my Q right?

You have to think about what's happening.

Is there enough heat absorption by the ice to cool the water to 0 ?

If not how much of the ice needs to be melted to bring it to 0 ?

If it all melts and there is still a surplus, then apply that surplus to the total water (water + melted ice because it's now water) and then figure the temperature of the whole.