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?

     q_metal = (m)(Dt)(cp)      q_water = (m)(Dt)(cp)

     q_{metal  =}  q_water

        (m)(Dt)(cp)    =     (m)(Dt)(cp)

      [1.0Kg][100C-T_f][0.50KJ]  =  [50.0g][T_f-20C][4.184 J]
                                   [   KgC]                                 [      gC ]

      [1.0Kg][1000g][100C-T_f][0.50KJ][   1Kg][1000J]  =  [50.0g][T_f-20C][4.184 J]
                   [   1Kg]                 [   KgC][1000g][   1KJ]                                [      gC ]

       [1000g][100C-T_f][0.50J]  =  [50.0g][T_f-20C][4.184 J]
                                    [   gC]                                [      gC ]

      [1000g][100C-T_f][0.50   J/gC]  = [T_f-20C]
       [ 50.0g]                [4.184 J/gC]

       [2.39][100C-T_f]  =  [T_f-20C]

       239C-2.39T_f  =  T_f-20C

       239C + 20C  =   T_f  +  2.39T_f  

       259C  =  3.39T_f

       259C  =  T_f
          3.39

        76C  =  T_f     

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?

     q_metal = (m)(Dt)(cp)      q_water = (m)(Dt)(cp)

     q_{metal  =}  q_water

        (m)(Dt)(cp)    =     (m)(Dt)(cp)

      [2.80Kg][100C-T_f][0.43KJ]  =  [50.0g][T_f-30C][4.184 J]
                                     [   KgC]                                 [      gC ]

      [2.8.0Kg][1000g][100C-T_f][0.43KJ][   1Kg][1000J]  =  [50.0g][T_f-30C][4.184 J]
                      [   1Kg]                 [   KgC][1000g][   1KJ]                                [      gC ]

       [2800g][100C-T_f][0.43J]  =  [50.0g][T_f-30C][4.184 J]
                                    [   gC]                                [      gC ]

      [2800g][100C-T_f][0.43   J/gC]  = [T_f-30C]
       [ 50.0g]                [4.184 J/gC]

       [24.08][100C-T_f]  =  [T_f-30C]

       2408C-24.08T_f  =  T_f-30C

       2408C + 30C  =   T_f  +  24.08T_f  

       2438C  =  25.08T_f

       2438C  =  T_f
          325.08

        97C  =  T_f    

 Answer: final temperature of metal and water is 97C