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INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY       93

             Worked Example 3.4 An electrical heater warms 12 g of water. Its initial temperature
                   ◦
             is 35.0 C. The heater emits 15 W for 1 min. What is the new temperature of the water?

             Answer strategy. Firstly, we will calculate the energy produced by the
             heater, in joules. Secondly, knowing the heat capacity of the water C,  The word ‘strategy’
             we divide this energy by C to obtain the temperature rise.   comes from the Greek
               (1) To calculate the energy produced by the heater. Remember that  stratos meaning ‘army’.
                                                      −1
                       −1
             1W = 1J s , so a wattage of 15 W means 15 J s . The heater oper-  Strategy originally con-
             ates for 1 min (i.e. 60 s), so the energy produced is 15 J s −1  × 60 s =  cerned military
             900 J.                                                       manoeuvres.
               This amount of energy is absorbed by 12 g of water, so the energy
             absorbed per gram is
                                           900 J        −1
                                                = 75 J g
                                           12 g

               (2) To calculate the temperature rise. The change in temperature  T is sufficiently small
             that we are justified in assuming that the value of C V is independent of temperature. This
             assumption allows us to employ the approximate equation, Equation (3.7). We rearrange
             it to make  T the subject:
                                                    U
                                              T =
                                                    C V
             Inserting values:
                                                 75 J g −1
                                          T =
                                                      −1 −1
                                               4.18 J K g
             yielding
                                             T = 17.9K

                                                                              ◦
                                                                     ◦
                 ◦
             As 1 C = 1 K, the final temperature of the water is (25.0 + 17.9) C = 42.9 C.
             SAQ 3.2 How much energy must be added to 1.35 kg of water in a pan if
                                                                           ◦
                                        ◦
             it is to be warmed from 20 C to its boiling temperature of 100 C? Assume
             C V does not vary from 4.18 J K −1  mol −1 .
               The heat capacity C V is an extensive quantity, so its value depends on how much
             of a material we want to warm up. As chemists, we usually want a value of C V
             expressed per mole of material. A molar heat capacity is an intensive quantity.



                                               Aside

                Another heat capacity is C p , the heat capacity measured at constant pressure (which is
                also called the isobaric heat capacity). The values of C p and C V will differ, by perhaps
                as much as 5–10 per cent. We will look at C p in more depth in the next section.
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