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[Heat and calorimeter] I can't get the correct answer

  1. Jun 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Hey there guys, first of all English is not my main language so it will be a little hard for me to say things in english. I am dutch native speaker so I will translate most of the things in english using google translator. I hope you guys are okay with that.

    Task :
    One wants the heat capacity of a calorimeter. It does so by one in the empty calorimeter, which has a temperature of 19.3 °C, 50 grams of benzene with a temperature of 13, 5 °C is added. This drops the temperature of the calorimeter to 15, 9 °C.
    a)Calculate the value of the heat capacity of this calormeter.
    b)Directly after this (the temperature of the calorimeter is still 15.9 °C), a Platinum bullet is added, which has a mass of 15.0 gram, in the calorimeter. This platinum ball has been kept in a glass flame beforehand; the temperature of the calorimeter runs to 23.7 °C . determine the temperature of the gas flame.

    Okay, I had no problem answering a.
    Now I did the same thing in b, but my answer does not match the one from my book. I tried to look if I missed any ##-## or ##+## sign , but still no luck. I even checked whether I filled the wrong number or not, and to me it seems that everything is correct. The Answer should be 604 °C , but I keep getting below 604 °C. I just want to know if I am doing something wrong or maybe the book answer could be wrong.

    2. Relevant equations
    ##Q = m.c. \bigtriangleup T##
    ##Q = C .ΔT##

    3. The attempt at a solution
    I've uploaded 2 pictures of the calculations I did.
    According to my book the answer I got for a is correct.
     

    Attached Files:

  2. jcsd
  3. Jun 17, 2017 #2
    I don't follow your algebra. Where did the 2 come from?
     
  4. Jun 17, 2017 #3
    ##15 × 0,13(Tend - Tbegin)##
    Tend is given which = 23,7 °C
    So it will be ##15 × 0,13(23,7 - Tbegin)##
    ##15 × 0,13 = 1,95##
    ##1,95(23,7 - Tbegin) = 46,22 - 1,95T##
     
  5. Jun 17, 2017 #4
    (468+667)/(1.95 )+ 23.7= ??
     
  6. Jun 17, 2017 #5
    Well the answer = 605,7 , but I'm still confused why you did ##+ 23,7## and not ##+46,22##
     
  7. Jun 17, 2017 #6
    I'm on my iPhone now, so it's hard to do equations. I'll get back with you later when I have my computer. But basically, the heat gained by the benzene and calorimeter is equal to the heat list by the bullet.
     
  8. Jun 17, 2017 #7
    Heat removed from bullet = ##1.95(T-23.7)##. So, $$468+667=1.95(T-23.7)$$
     
  9. Jun 17, 2017 #8
    Ah that explains it, thank you sir for your help. :smile:
     
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