Question

A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Answer 1

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is  $T_i$ = 0°C

Temperature of  steam  $T_s$ = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          $Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion  and the value is $334*10^3 J/kg$

           $Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

            $Q_B = M_s * L_v$

           Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

           $Q_B = 0.019 * 2256*10^3 = 42864J$

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 $Q_C = M_s *C_water ($100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

                  $Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

             $Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        $Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

                                               $T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

                                               $T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

                                              $T = 12.48^oC$