Calculating Specific Heat of Alcohol Using a Copper Calorimeter

[kriegera]

Homework Statement

A mass of 200 grams of copper, whose specific heat is 0.095, is heated to 100° C, and placed in 100 grams of alcohol at 8° C contained in a copper calorimeter, whose mass is 25 grams, and the temperature rises 28.5°C. Find the specific heat of the alcohol.

Homework Equations

ΔH=cmΔT
C=H/mΔT

The Attempt at a Solution

The heat of placing the copper into the calorimeter is the energy used to warm the alcohol via conduction. We first have to decide how much heat it takes to raise the temp of the alcohol from 8° to 28.5°.
Use equation:
ΔH=cmΔT = (0.095)(200g)(100-0)
Since no beginning heat is given, we will assume the copper was heated from 0°C.
= 1,900 J

Knowing the force of heat energy that heats the alcohol, we can then rearrange the equation to find the specific heat of alcohol:
C=H/mΔT = 1,900/(100g)(28.5-8) = 0.93 J/g.K


Am i on track here?

 

[rl.bhat]

Heat lost by the copper = mc*cc*(T1 - T) where T is the final temperature of the mixture.
Heat gained by the alcohol along with calorimeter is
(mc*cc + ma*ca)(T - T2) where T2 is the initial temperature of the alcohol.

 

[kriegera]

i understand now we need to find the the heat gained by the calorimeter to find the total energy for the final part of the equation but there are still components missing.

1)how much heat is lost by the copper:
ΔH=cmΔT = (0.095)(200g)(100-28.5).
= 1,358.5 J

2)how much heat is gained by alcohol and calorimeter
(25)(cc) + (100)(ca) (28.5-8)

Then Ca = Heat Gained - Heat Lost/mΔT

but how do you find cc and ca for part 2?

 

[rl.bhat]

In [(25)(cc) + (100)(ca)] (28.5-8)
cc is the specific heat of the calorimeter which is given. And ca is the specific heat of alcohol which you have to find out.

 

[kriegera]

So we don't have to do this: Then Ca = Heat Gained - Heat Lost/mΔT first right?

In In [(25)(cc) + (100)(ca)] (28.5-8)
i dont' think the specific heat of the calorimeter is given - the specific heat of the copper is given. Would you use:

(25)(0.092) + (100)(CA)(20.8) = 0

 

[rl.bhat]

(0.095)(200g)(100-28.5) = [(25)(0.095) + (100)(CA)](20.8)

Note the bracket.

 

[kriegera]

almost there i think
1358.5 = [2.375 + (100)(ca)] (20.8). i ge tthis but if you solve this for ca=.629
We're looking for the "heat gained" value not ca in this equation. how do you find the heat gained value from the above equation to solve the one below?
[Ca = heat Gained - Heat Lost]/mΔT first right

 

[rl.bhat]

"We're looking for the "heat gained" value not ca in this equation."
Heat gained by the alcohol and calorimeter is equal to heat lost be the water.
And in the problem they have asked to find the specific heat of the alcohol.

 

[kriegera]

If heat gained is equal to heat lost then the value would be 1358.5.
When you set the equation like this don't you have to solve for ca? it's the only unkown.
1358.5 = [2.375 + (100)(ca)] (20.8)

and if heat gained is equal to heat lost then the numerator of our equation will be 0 right?
Ca = heat Gained - Heat Lost]/mΔT first right

 

[rl.bhat]

"Ca = heat Gained - Heat Lost]/mΔT"
From where did you get the above expression? This is wrong.

 

[kriegera]

Oh i thought that was the final step. So if heat gained equaled heat lost then heat gained would be 1358.5 J. And
1358.5 = [2.375 + (100)(ca)] (20.8)
Ca = .629 ?

 

[rl.bhat]

Yes.

 

[kriegera]

Thank you! :)