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Finding specific heat with unknown Q

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data Hello, I'm having some trouble with a lab report. I am supposed to find the specific heat of five metals but I don't know exactly how to use the equation correctly. for instance I have copper at 70.2g, the delta T at 1.8 degrees Celsius, specific heat of copper is 385 J/kg-degrees Celsius.

    2. Relevant equations
    Q=MC*delta T
    C=Q/M*delta T

    3. The attempt at a solution
    I used the standard value of copper's specific heat to find Q then plug Q back into the second equation to find our lab's value of copper's specific heat. This however gives me the same answer of course. My teacher wants us to use our data to come up with coppers specific heat and compare it to the actual specific heat but I can't figure out how to get a value for Q without using coppers standard specific heat. Any help would be greatly appreciated.
  2. jcsd
  3. Jan 17, 2010 #2
    Are you sure you are understanding the assignment?

    Specific heat capacity is originally determined by mechanical means.
  4. Jan 17, 2010 #3
    I'm pretty sure. We were supposed to use our findings to calculate the specific heat of the metals we tested. Yet I can't figure out how to find another specific heat. We are supposed to find the specific heat then compare it to the actual established specific heat of the metal and determine the relative error of our lab in each metal.
  5. Jan 17, 2010 #4
    So you actually tested the metals, and accumulated the data?

    If so, can you describe the test?
  6. Jan 17, 2010 #5
    We weighed the 5 metals. We weighed the calorimeter cups with and without water. We heated a beaker of water with the metals in them to boiling. We placed the metals in the calorimeter cup one at a time until each reached thermal equilibrium. then took down the temperature. That's the extent of it. I talked to a friend in the class and he said Q was equal to the temperature of the metals before we brought them over to the calorimeter cups. so 103 Celsius= the mass* specific heat* the temp change.
  7. Jan 17, 2010 #6
    What you measured with the calorimeter was Q.

    The difference in temperature from before and after.

    Convert calories to joules. Then plug that value into your equation.
  8. Jan 17, 2010 #7
    So the change in temperature was Q? for instance one calorimeter started at 18.5 Celsius and we added the copper to it and it reached thermal equilibrium at about 22.1 Celsius so Q would be the change between those two? I'm a bit lost as to converting calories to joules because forgive my ignorance but we measured in celsius i don't know how to get that in calories.
  9. Jan 17, 2010 #8
    A calorie is the amount of heat needed to raise the temperature of 1 gram of water 1 degree Celsius.

    3.6 times mass of the water will give you the calories. 4.184 joules equals 1 calorie.

    You should be able to take it from there.
  10. Jan 17, 2010 #9
    Alright So I worked out the problem how you said.
    mass of water is 242.3*3.4 which gives me 823.82 calories, multiplied by 4.184 which gives me 3445 joules. I plugged that into Q=mc delta T, 3445J=.07kg*C*1.8 celsius, which resulted in 27341 J/kg-degree celsius. I just want to be sure I'm doing this right.
  11. Jan 17, 2010 #10
    What was the temperature of the copper before and after?
  12. Jan 17, 2010 #11
    The temperature of the copper before being put in the calorimeter was 103 Celsius. I believe we were to assume the water and metals to be the same temperature. The water in the calorimeter was 18.5 Celsius and 21 Celsius after thermal equilibrium.
  13. Jan 17, 2010 #12

    So the delta temperature is 82C.

    Q=3649.6j ((3.6*242.3)*4.184)



    At least, that is how I understand the equation.
  14. Jan 17, 2010 #13
    hmm, the .063 is the specific heat of the copper sample? I don't know, I think we are thinking of two different concepts or whatnot. Thanks for all your help, I do appreciate it:approve:
  15. Jan 17, 2010 #14
    You need to get your numbers right.

    22.1 or 21?






    That is a little more realistic.
  16. Jan 17, 2010 #15
    Alright, thanks for all your help, God bless you :smile:
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