Calorimetry Lab: Calculating Specific Heat Capacity

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To calculate the specific heat capacity of the unknown material in the calorimetry lab, it is essential to consider the heat transfer between the sample, the calorimeter can, and the water separately due to their differing specific heats. The problem involves a 0.0850-kg sample at 100 degrees C, a 0.150-kg copper calorimeter, and 0.200 kg of water, with a final temperature of 26.1 degrees C. The correct approach is to set up three heat equations: one for the sample, one for the copper can, and one for the water, ensuring that the heat lost by the sample equals the heat gained by the calorimeter and water. By calculating the individual heat exchanges, the specific heat capacity of the unknown material can be determined. This method effectively accounts for all components involved in the calorimetry process.
cukitas2001
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Hey I am stuck on yet another one. This one has me stumpted becasue it has to do with a caliometry can and I am given mass of the can and water in it but i don't know if i should add them or have them in separate m*c*T expressions. Anywho here's the problemo:

A laboratory technician drops a 0.0850-kg sample of unknown material, at a temperature of 100 degrees C, into a calorimeter. The calorimeter can, initially at 19.0 degrees C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can is 26.1 degrees C. Compute the specific heat capacity of the sample.

What i was thinking was along the lines of Qsample+Qcaliometer but I am not taking into account the copper mass or the water mass...how can i?
 
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They are separate because they have differing specific heats.
 
So then i should have three Q's?
 
Yes, that's right.
 
Ok got it...thank u tons
 

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