Question

A pot of hot water is boiling on your stove.  The pot is in the shape of a round cylinder made of aluminum which is 10 cm in radius.  The aluminum is 3 mm thick.  You notice that after 22 minutes, 0.13 kg of water has boiled away.

A. What is the temperature of the inner surface of the bottom of the pot?

B. How much energy is required to boil away M kg of water?

C. What is the difference in temperature between the inner and outer sufaces of the aluminum bottom of the pot?

D. What is the change in entropy of the boiling water after boiling away 0.13 kg of water?

Answer 1

PART A    The temp of the inner surface of the bottom of the pan has to be the boiling temperature of the water, which is 100 deg C.

PART B      The energy to boil away (ie, to convert from liquid to steam phase) is

                    

                              (mass of water)(Latent heat of vaporization of water)

      

               (M kg) ( 2.26 x 10^6 J/kg) =    2.26 x 10^6 M   (units will be Joules for the energy).  

PART C       Heat energy required to vaporize 0.13kg of water is (0.13 kg)(2.26 x 10^6 J/kg) = 293,800 J

Heat energy per sec flowing into pan is    293,800J / [(22 min)(60sec/min)]    =   222.6 J/sec.  

   The heat flow eqn is     Q / t =   k A (delta T) / (thickness),   where Q/t = 222.6 J/sec,

k = thermal conductivity of water, A is cross sectional area of the cylindrical pan, and delta T is the Temp difference between inner and outer surfaces of bottom of pan .

   k = 0.6 J / sec-m-degC          A = pi R^2         thickness = 3 x 10^ -3 m

              222.6 =   (0.6) (pi) (0.10m) ^2 ( delta T) / ( 3 x 10^ -3 m )      solve this for (delta T) to get temp difference:

   delta T =    35.4 deg C

PART D      Change in Entropy is defined as   delta Q / T for a constant temp, which we have when the water is boiling and changing phase.

                 change in entropy =   293,800 J / (100+273) = 787.7   J/deg K      this is the increase in entropy of the water.