Page 152 - text book form physics kssm 2020
P. 152

Solving Problems Involving Latent Heat


                   Example 1

                  Figure 4.22 shows a 480 W immersion heater          To power
                  used to melt ice in a container. In 120 s, the      supply              Immersion heater
                  reading of the electronic balance decreases by
                  0.172 kg.
                                                                                          Crushed ice
                  (a)  What is the mass of ice that has melted
                      during the heating period?
                  (b)  Calculate the specifi c latent heat of fusion of   Electronic  ON
                                                                            OFF
                      ice, l .                                  balance                         To sink
                          f
                                                                              Figure 4.22
                   Solution:
                  (a) Make assumptions:
                      (i)  Ice is melted by heat from the immersion heater only.
                      (ii)  All water from the melting ice fl ows out of the container.
                      Relate the change in the readings of the electronic balance to the mass of ice which has
                      melted:
                      Mass of ice melted = Decrease in the reading of the electronic balance
                                     m = 0.172 kg
                  (b) Make assumptions:
                      (i)  All heat supplied by the immersion heater is absorbed by the melting ice.
                      (ii)  No transfer of heat from the surrounding into the apparatus.


                   Step                             m = 0.172 kg
                   List the given information     14243 P  = 480 W
                   in symbols.                      t  = 120 s


                   Step
                   Identify and write down the     123 Pt = ml f
                   formula used.

                   Step                              480 × 120 = 0.172 × l f
                   Substitute numerical values into         l  =   480 × 120
                   the formula and perform the    14243      f    0.172
                                                                        5
                   calculations.                              = 3.35 × 10  J kg −1












                  146                                                                              4.3.5
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