Magicarp
Jun14-09, 04:19 PM
1. The problem statement, all variables and given/known data
An insulated beaker with negligible mass contains liquid water with a mass of 0.275kg and a temperature of 77.7.
How much ice at a temperature of -17.4 must be dropped into the water so that the final temperature of the system will be 29.0?
Take the specific heat of liquid water to be 4190 , the specific heat of ice to be 2100 , and the heat of fusion for water to be 334 x 10^3 .
2. Relevant equations
Q = mc\DeltaT
Q_tot = q_1 + q_2 ..etc
Fusion - Phase change
3. The attempt at a solution
I am almost certain I have done this correctly, but masteringphysics will not admit it! :/
I will leave off some units as I have made sure they all match up.
q1: Ice to 0C
q2: Ice fusion mw= mass water
q3: Ice to 29C mi = mass ice
q4: H20 to 29C
((mw)(4190)(29C-77.7C) + (mi)(2100)(0C+17.4C)+(mi)(334 X 10^3)+(mi)(4190)(29C-0C))= 0
I solve for mi and get 0.114kg ice. MP no likey. Anyone care to check this?
Thanks in advance,
Magicarp
An insulated beaker with negligible mass contains liquid water with a mass of 0.275kg and a temperature of 77.7.
How much ice at a temperature of -17.4 must be dropped into the water so that the final temperature of the system will be 29.0?
Take the specific heat of liquid water to be 4190 , the specific heat of ice to be 2100 , and the heat of fusion for water to be 334 x 10^3 .
2. Relevant equations
Q = mc\DeltaT
Q_tot = q_1 + q_2 ..etc
Fusion - Phase change
3. The attempt at a solution
I am almost certain I have done this correctly, but masteringphysics will not admit it! :/
I will leave off some units as I have made sure they all match up.
q1: Ice to 0C
q2: Ice fusion mw= mass water
q3: Ice to 29C mi = mass ice
q4: H20 to 29C
((mw)(4190)(29C-77.7C) + (mi)(2100)(0C+17.4C)+(mi)(334 X 10^3)+(mi)(4190)(29C-0C))= 0
I solve for mi and get 0.114kg ice. MP no likey. Anyone care to check this?
Thanks in advance,
Magicarp