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Exercise Solution

2.    A Kcal is defined as the heat needed to raise 1 kg of water by 1 Co.   We use that definition to find the temperature change.  The final temperature can then be found.

 

      (7700 J/3.0 kg)(1 kcal/4186 J)(1 kg)1C)/1 kcal = 0.61 C    so the final temperature is this added to original = 10.6 C

 

9.    The specific heat can be calculated from Eq. 14-2.

 

       Q = mcDT  ð c = Q/mDT = 1.35x105J/(5.1 kg)(31.5 C - 18.0 C) = 1961 J/kg C  = 2.0 x103 J/kgC

 

15.  The heat must warm both the water and the pot to 100oC.  The heat is also the power times the time.

 

       Q = Pt = (mAlcAl + mH20CH20)DTH20    ð about 7 minutes or 425 seconds

 

22.  Assume that the heat from the person is only used to evaporate the water.  Also, we use the heat of vaporization at room temperature (585 kcal/kg), since the person’s

        temperature is closer to room temperature than 100oC.

      Q = mLvap ð m = Q/Lvap = 180 kcal585 kgal/kg = about .31 kg = 310 ml 

34.  The heat conduction rate is given by Eq. 14-4.

 

        Q/t = kA(T2 - T1)/l = (0.84 J/smC)(3.0 m2)[15.0 C - (-5 C)]/3.2x10-3 m  = 1.6 x 104 W           

 

35b.  The net energy flow rate is given by equation 14-6 with a temperature of the surroundings = 268 K

                    

         DQ/Dt = esA(T24 - T14) = 90.35)(5.67x10-8W/M2k4)4p(0.22 m)2 [(298 K)4 - (268 K)4] = 33 W