The Final Temperature and Remaining Ice in a Calorimetry Experiment

[olso4142]

[SOLVED] phase changes

Homework Statement

A 40.0 g block of ice is cooled to -78oC. It is added to 560 g of water in an 80.0 g copper calorimeter at a temperature of 25oC. Determine the final temperature . If not all the ice melts, determine how much ice is left.

Homework Equations

Q=mC∆T
Q=mL (phase change)
Q(lost)+ Q(gained)=0

The Attempt at a Solution

energy to bring water to O degrees C
Q=mC∆T
=(.040kg)(4186J/KgC)(0-25C)
=-4186J

Energy to bring copper to 0 degrees C
Q=mC∆T
=(0.800kg)(386J/kgC)(0-25C)
=-7720

Energy to bring ice to 0
Q=mC∆T
=(0.040kg)(?)(0--78)
=? is the specific heat of ice the same as water?

energy phase change
Q=mL
=(.040kg)(?)
is the latent heat of ice the same as the latent heat for freezing water?

and then after finding all that how do i know if there is ice left?

 

[marcusl]

Homework Statement

A 40.0 g block of ice is cooled to -78oC. It is added to 560 g of water in an 80.0 g copper calorimeter at a temperature of 25oC. Determine the final temperature . If not all the ice melts, determine how much ice is left.

Homework Equations

Q=mC∆T
Q=mL (phase change)
Q(lost)+ Q(gained)=0

The Attempt at a Solution

energy to bring water to O degrees C
Q=mC∆T
=(.040kg)(4186J/KgC)(0-25C)
=-4186J

Not quite. You need to use the mass of the water here.

Energy to bring copper to 0 degrees C
Q=mC∆T
=(0.800kg)(386J/kgC)(0-25C)
=-7720

The copper mass is 0.08 kg

Energy to bring ice to 0
Q=mC∆T
=(0.040kg)(?)(0--78)
=? is the specific heat of ice the same as water?

energy phase change
Q=mL
=(.040kg)(?)
is the latent heat of ice the same as the latent heat for freezing water?

and then after finding all that how do i know if there is ice left?

Specific heat of ice is temperature dependent. Using value 1820 J/kg-C (close to the average temperature of -40C) should give good results.

Latent heat is defined as energy difference between ice and water and 0C, so the answer to your second to last question is "yes".

To solve the problem, compare the total energy content of copper + water (Cu+W) to the energy it takes to melt the ice. If E(Cu+W) is lower, than work the ice calculation backwards to find the mass that will melt. At the end, of course, everything will be at 0C.

If E(Cu+W) is higher, then it loses the energy to melt the ice. Calculate the temperature T of the Cu+W after ice is melted. Now you have Cu+W at T and .04kg of water at 0C so you can find the final temperature from that.

 

[olso4142]

Can i just add Qw+QI+Qc and then since that is greater than Qphase all of the ice melts. does that work?