Page 29 - FINAL DBS10012_HEAT TEMPERATURE 20102022 (2) (1)
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2. How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C?

                           [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water =
                           4200J/ (kg K).]


                          Solution:


                               m = 2kg
                               Specific latent heat of fusion of water, L = 334 000 J/kg

                               specific heat capacity of water = 4200 J/ (kg K)



                      3. Starting at 20°C, how much heat is required to heat 0.3 kg of aluminum to its melting point and
                      then to convert it all to liquid? [Specific heat capacity of aluminium = 900J kg-1 °C-1; Specific

                      latent heat of aluminium = 321,000 Jkg-1, Melting point of aluminium = 660°C]


                      Solution:
                                                   T,  C
                                                     o
                                Energy needed to melt 2kg of ice,    Q = mL
                                                 660 C
                                                    o
                                Q1 = mL = (2) (334000) = 668000J

                                                    o       Q = mcθ
                                                  20 C
                                Energy needed to change the temperature from 0°C to 20°C.

                                Q2 = mcθ = (2) (4200) (20 - 0) = 168000J   Heat energy supplied
                                Total energy needed = Q1 + Q2 = 668000 + 168000 = 836000J
                            m = 0.3kg

                            L aluminium = 321 000 J/kg
                            c aluminium = 900 J/ (kg K)

                            Energy needed to increase the temperature from 20°C to 660°C


                            Q1 = mcθ = (0.3) (900) (660 - 20) = 172,800J
                            Energy needed to melt 0.3kg of aluminium,


                            Q2 = mL = (0.3) (321000) = 96,300J


                            Total energy needed = Q1 + Q2 = 172,800 + 96,300 = 269,100J


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