Finding end temperature of mixed substances

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To find the end temperature of a mixture involving water, glycerol, and a calorimeter, the conservation of energy principle is applied. The energy gained by the water and calorimeter as they warm up must equal the energy lost by the glycerol as it cools down. The specific heat capacities for each substance are provided, and the initial temperatures are noted. A common mistake involves incorrect placement of temperature variables in the equations. After correcting the formulation, the problem can be solved accurately, yielding a final temperature between 10°C and 20°C.
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hi I am 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 I am unsure of how to formulate the equation.
 
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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:
dthmnstr said:
hi I am 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 I am unsure of how to formulate the equation.

Did you remember to convert to kelvin?
 
thanks dudes for the help, i just messed up with placing the T in the equation so i got it solved properly now
 
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