How Does a Calorimeter's Composition Impact Heat Transfer Calculations?

[mirandasatterley]

"An aluminum calorimeter with a mass of 100g contains 250g of water. The calorimeter and water are in thermal equilibrium at 10.0Degree C. Two metalic blocks are placed into the water. One is a 50g piece of copper at 80.0 degree C. The other block has a mass of 70.0g and is origionally at a temperature of 100degree C. the entire system stabalizes at a final temerature of 20degree C. Determine the specific heat of the unknown sample.

I know that for general cases;
Q(water) = -Q(x) Where x is the substance put into the water
mcdeltaT(water) = - mcdeltaT(x)

My first question is: Does the fact that the calorimeter is made of aluminum affect the Q(calorimeter) part, which is Q(water) in my previous equation? If it does, any hints on how. I think I have to add the masses, but I am unsure about the specific heat part.

My other question is: How do I deal with two objects being put into the water?

Any help is appreciated.

 

[OlderDan]

The aluminum warms up with the rest of the system. You will need to include it in your heat transferr calculations.

 

[mirandasatterley]

"An aluminum calorimeter with a mass of 100g contains 250g of water. The calorimeter and water are in thermal equilibrium at 10.0Degree C. Two metalic blocks are placed into the water. One is a 50g piece of copper at 80.0 degree C. The other block has a mass of 70.0g and is origionally at a temperature of 100degree C. the entire system stabalizes at a final temerature of 20degree C. Determine the specific heat of the unknown sample.

How do I deal with two objects being put into the water?QUOTE]

 

[OlderDan]

How do I deal with two objects being put into the water?

The water, copper, and aluminum have known masses and specific heats (you may have to look them up). The only unknown is the specific heat of the unknown sample.

 

[mirandasatterley]

The water, copper, and aluminum have known masses and specific heats (you may have to look them up). The only unknown is the specific heat of the unknown sample.

Can i use the equation;

Q(calorimeter) + Q(water) = Q(Block 1) + Q(Block 2)

Then solve for the specific heat capacity of block 2?
Is it correct to add them like This?

 

[Amith2006]

Can i use the equation;

Q(calorimeter) + Q(water) = Q(Block 1) + Q(Block 2)

Then solve for the specific heat capacity of block 2?
Is it correct to add them like This?

Yes. Since the final temperature attained is 20 deg C, the calorimeter and water gains heat energy while the 2 blocks lose energy.