# Finding end temperature of mixed substances

1. Jan 17, 2010

### dthmnstr

hi im having a bit of a problem with this anybody care to help me out?

a calorimeter contains 0.761kg of water. both calorimeter and water have the temperature 4 degrees celsius. 0.054kg of liquid glycerol at 26 degrees celsius is then poured into the calorimeter. find the end temperature of the entire mixture.

C(calorimeter)=86J/kgK C(water)=4180J/kgK C(glycerol)=2430J/kgK

so far i've tried the equations Q(calorimeter)+Q(water)=Q(glyserol) and Q(calorimeter)+Q(water)=Q(glycerol)+Q(glycerol turning solid) but i always get some kind of brutal answer, i think the answer lies somewhere between 10 'c and 20 'c but im unsure of how to formulate the equation.

2. Jan 17, 2010

### Stonebridge

Use the first equation you tried.
Let the temperature of the final mixture be, t, for example.
Find the energy gained by the water in rising to that temperature from 4 degrees.
Find the energy gained by the calorimeter in rising to that temperature from 4 degrees
Find the energy lost by the glycerol in falling to that temperature from 26 degrees
Use conservation of energy. Energy gained by Water and calorimeter = energy lost by glycerol. (Assuming no loss to surroundings) Solve for t.

Last edited: Jan 17, 2010
3. Jan 17, 2010

### Happyzor

Did you remember to convert to kelvin?

4. Jan 17, 2010

### dthmnstr

thanks dudes for the help, i just messed up with placing the T in the equation so i got it solved properly now