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

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To lower the temperature of 125 g of water from 23°C to 12°C, the energy required is calculated as -5760 J. This energy must be balanced by the heat absorbed from the melting ice, which has a latent heat of 3.34 x 10^5 J/kg. The initial calculation suggests that 17.2 g of ice is needed, but the correct answer is actually 15 g. The discrepancy arises because the melting ice will become water and mix with the existing water, which must be accounted for in the calculations. Proper consideration of the mixing process is essential for accurate results.
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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?
 
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Hi veronicak5678! :smile:
veronicak5678 said:
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! :wink:
 
I'm sorry, but I don't follow...
Should I be using something other than latent heat?
 
yes … the ice will become water, and mix with the water already there :smile:
 
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