A house is maintained at a comfortable temperature by means of an electrical resistor heater during winter. The heater is operated at a constant current (I) under an applied voltage of 110 V. For simplicity, let us assume that the heater operates at a steady state, with 100% of conversion efficiency from the electrical energy to the internal energy of air in the house. (a) (5p) Find I if the homeowner pays $8.80/day for heating, with electricity cost $0.09/kWh. (b) (5p) Calculate the heat produced by the heater in an hour.

a)   I = 0.01029 A , b)   Q = 4074.05 J

Explanation:

a) Let's work in parts, let's start by looking for the energy consumed.

For this we use a rule of three or direct proportions if 1 kWh costs $ 0.09, how much kWh is $ 8.80

           Energy = 8.80 / 0.09

           Energy = 97,778 kWh

to calculate the current, we must have the energy consumed per second

Let's use a rule of three if 97,778 kWh were consumed in 24 h = 86,400 s how much was consumed in 1 s

            E₀ = 97.778 1/86400

            E₀ = 1.13168 Ws

the power consumed is

           P = E₀ / t

           P = 1.13168 / 1

           P = 1.13168 W

power is

           P = I V

           I = P / V

           I = 1.13168 / 110

           I = 0.01029 A

b) the heat is given by

as the efficiency is 100%

          E = P t = Q

          Q = 1.131368 3600

          Q = 4074.05 J

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.