Finding end temperature of mixed substances

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Homework Help Overview

The problem involves a calorimeter containing water and glycerol, where the goal is to determine the final temperature of the mixture after the glycerol is added. The specific heat capacities of the substances involved are provided, and the initial temperatures are noted.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss using conservation of energy principles, specifically the energy gained by the water and calorimeter compared to the energy lost by the glycerol. There are attempts to formulate the equations needed to solve for the final temperature.

Discussion Status

Some participants have offered guidance on how to set up the equations, suggesting the use of a variable for the final temperature and emphasizing the conservation of energy. There is acknowledgment of a mistake in the setup of the equations by one participant, which has since been resolved.

Contextual Notes

One participant raised a question about the necessity of converting temperatures to Kelvin, indicating a potential assumption or detail that may need clarification in the context of the calculations.

dthmnstr
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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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