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Laboratory technician drops a 0.0850kg sample of unknown solid

  1. A laboratory technician drops a 0.0850kg sample of unknown solid material, at a temperature of 100.0 degree celcius, into a calorimeter. The calorimeter can, initially at 19.0 degrees celcius, is made of 0.150kg of copper and contains 0.200kg of water. The final temperature of the calorimeter can and contents is 26.1 degrees celcius. Compute the specific heat of the sample.

    The only thing i have so far is

    Qsys= -Qsurr
    but im stumped with the 2 specific heat variables
  2. jcsd
  3. Borek

    Staff: Mentor

    Re: Thermodynamics

    Not sys and surr, I would rather go for gain=lost.

    I suppose you should check both specific heats of copper and water in tables.

  4. epenguin

    epenguin 2,218
    Homework Helper
    Gold Member

    Re: Thermodynamics

    Being calories and degrees C he ought to know that for water roughly.
  5. Re: Thermodynamics

    i'll be so grateful if you can help me for this problem:
    a 30.14-g stainless steel ball bearing at 117.82 c is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44 C.if the specific heat of the ball bearing is 0.474 J/g.c, calculate the final temperature of the water.assume the calorimeter to have negligible capaity.
  6. Borek

    Staff: Mentor

    Re: Thermodynamics

    It is again simple heat gain (by water) equals heat lost (by ball). Assume final temperature to be Tfinal and write equation for a heat balance. You will get equation with one unknown. That's all.

  7. Re: Thermodynamics

    thank you so much,,i got the idea, but i'm still thinking about if q(water) equals q(ball), can we
    prove that (Delta Temp) for water equals (Delata Temp) for the ball!!!!!
  8. Re: Thermodynamics

    No, but both changes in temperature are directly proportional to the amount of heat transferred.
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