Question

A family comes home from a long vacation with laundry to do and showers to take. The water heater has been turned off during vacation. If the water heater has a capacity of 49.1 gallons and a 4610 W heating element, how much time is required to raise the temperature of the water from 18.8°C to 62.1°C? Assume that the heater is well insulated and no water is withdrawn from the tank during this time.

Answer 1

specific heat capacity of water C = 4186 J/kg/^oC

             capacity of water in heater = 49.1 gallons

                                                      = 49.1 x 3.785 = 185.84 litre

                     185 .84 litre of water = 185.84 kg

Heat required to raise the temperature from 18.8^oC to 62.1^oC

                    H = 4186 x ( 62.1 - 18.8 ) x 185.84

                        H = 33684206.19 J

power of heater P = 4610 W                                       1 W = 1 J/s

                             t = 33684206.19 / 4610

                             t =7306.8 s

                              t = 7306.8 / 60 x 60

                             t = 121.7 min = 2 hr 2 min

                          t = 2hr 2 min