Physics Question?

Question

Mr Smith has a 3kW. water heater containing 120kg. of water initially as 20°C, and a clock radio which consumes 5W. No other appliances are using any electricity. There is 41p. of credit remaining on his electricity meter.

The alarm is set to sound after 8 hours, but if the meter runs out sooner than that then Mr Smith will oversleep.

What is the hottest temperature that his thermostat could be set to without running the electricity meter out of credit?

Specific Heat Capacity of water = 4200 J kg-1 K-1.
Price of electricity = 8p/kWh.
Assume the water heater is thermally insulated, and neglect its own heat capacity.

Answer 1

Let T = the maximum temperature to which the thermostat is set, then the temperature of water should not reach this value in 8 hours.
In 41 p, amount of electricity available = 41/8 = 5.125 kWh
Of this, amount of electricity consumed by clock = 5 x 8 = 40 Wh
Balance, available for consumption by water heater = 5.085 kWh = (5.085) * (3600000) J
This should equal the energy for heating 120 kg of water
=> 120*(T - 20)*(4200) = (5.085) * ( 3600000)
=> T - 20 = (5.085) * (3600) / [(12) * (42)]
=> T - 20 = 36.32
=> T = 56.32° C.