Latent Heat Q: How to Lower Temp of 125 g Water to 12°C

[veronicak5678]

Homework Statement

A well insulated cup contains 125 g of water at 23 degrees. How many grams of ice must be added to lower temp to 12?

Homework Equations

Q = c*m*delta t

The Attempt at a Solution

energy needed to lower temp:
Q = 4190 J / kg * (-11 degrees) * (.125 kg) = -5760 J

Q from ice = - Q from water = 5760 J

to find mass of ice needed:

5760 J = Latent heat of ice * mass needed = (3.34 * 10^5 J / kg) * m

= .0172 kg = 17.2 g

The correct answer is 15 g. What am I doing wrong?

 

[tiny-tim]

Hi veronicak5678!

A well insulated cup contains 125 g of water at 23 degrees. How many grams of ice must be added to lower temp to 12?
…
The correct answer is 15 g. What am I doing wrong?

hmm … you didn't stir!

 

[veronicak5678]

I'm sorry, but I don't follow...
Should I be using something other than latent heat?

 

[tiny-tim]

yes … the ice will become water, and mix with the water already there