Saturday, October 16, 2010

SPECIFIC HEAT PROBLEMS

1. What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0C to 70.0C?


q = (m)(Dt)(cp)   where q  is the energy in joules
                                          m is the mass
                                         
Dt is the change in initial temperature to the final temperature
                                          cp is the specific heat (heat required to raise 1 g of the substance 
1C)

q = (m)(Dt)(cp)
     2.5 x 103 J = (1.0 x 104 g)(70C-10C)(cp)

    [2.5 x 103 J]                 = cp
     [1.0 x 104 g][70C-10C]
   cp=   0.0042 J   
                  gC
Answer:  cp = 0.0042 J/gC




2) A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a  50.0 g sample of water at 20.0C.  What is the final temperature of the metal and the water?
     qmetal = (m)(Dt)(cp)      qwater = (m)(Dt)(cp)
     qmetal  =  qwater
        (m)(Dt)(cp)    =     (m)(Dt)(cp)
      [1.0Kg][100C-Tf][0.50KJ]  =  [50.0g][Tf-20C][4.184 J]
                                   [   KgC]                                 [      gC ]
      [1.0Kg][1000g][100C-Tf][0.50KJ][   1Kg][1000J]  =  [50.0g][Tf-20C][4.184 J]
                   [   1Kg]                 [   KgC][1000g][   1KJ]                                [      gC ]
       [1000g][100C-Tf][0.50J]  =  [50.0g][Tf-20C][4.184 J]
                                    [   gC]                                [      gC ]
      [1000g][100C-Tf][0.50   J/gC]  = [Tf-20C]
       [ 50.0g]                [4.184 J/gC]
       [2.39][100C-Tf]  =  [Tf-20C]
       239C-2.39Tf  =  Tf-20C
       239C + 20C  =   T+  2.39Tf  
       259C  =  3.39Tf
       259C  =  Tf
         
3.39
        76C  =  Tf     
Answer: Final temperature of metal and water is 76C


3)  3) A 2.8 kg sample of a metal with a specific heat of 0.43KJ/KgC is heated to 100.0C then placed in a 50.0 g sample of water at 30.0C.  What is the final temperature of the metal and the water?

     qmetal = (m)(Dt)(cp)      qwater = (m)(Dt)(cp)
     qmetal  =  qwater
        (m)(Dt)(cp)    =     (m)(Dt)(cp)
      [2.80Kg][100C-Tf][0.43KJ]  =  [50.0g][Tf-30C][4.184 J]
                                     [   KgC]                                 [      gC ]
      [2.8.0Kg][1000g][100C-Tf][0.43KJ][   1Kg][1000J]  =  [50.0g][Tf-30C][4.184 J]
                      [   1Kg]                 [   KgC][1000g][   1KJ]                                [      gC ]
       [2800g][100C-Tf][0.43J]  =  [50.0g][Tf-30C][4.184 J]
                                    [   gC]                                [      gC ]
      [2800g][100C-Tf][0.43   J/gC]  = [Tf-30C]
       [ 50.0g]                [4.184 J/gC]
       [24.08][100C-Tf]  =  [Tf-30C]
       2408C-24.08Tf  =  Tf-30C
       2408C + 30C  =   T+  24.08Tf  
       2438C  =  25.08Tf
       2438C  =  Tf
         
325.08
        97C  =  Tf    
 Answer: final temperature of metal and water is 97C

2 comments:

  1. the solutions to problems can be simply understood.. :)

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  2. The blog is very useful and lucid indeed but you could have given the values of the problem in a simpler manner.

    ReplyDelete