The heating coil of a hot-water heater has a resistance of 20 and operates at 210 V. If electrical energy costs $0.080/kWh, what does it cost to raise 270 kg of water in the tank from 15°C to 72°C?
2007-02-15 15:39:53 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
We need to know the specific heat of water (the amount of energy to raise 1mL by 1 degree C). I shall call it 's'.
Energy to heat water = 270kg x 1000ml/Kg x (72-15) x s
= 1.539x10^7 s Joules
Total Cost = Energy X Cost per unit energy
= 1.539x10^7 x s x 0.08 $/kWh / 1000 w/Kw / 3600 seconds/hour
Note that the voltage and resistance don't come into the calculation. They would only be used to calculate the time that the heating would take.
2007-02-15 16:02:28 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Making a *lot* of assumptions ☺
The energy given off by the heater coil is
V²/R = 210²/20 = 2205 KW = 2205 J/s
The heat capacity of water is 4.186 J/g-degree so
4186 J/kg-degree and the delta-t is
72 - 15 = 57 degrees. So
4186*270*57 = 64422540 J supplied at 2205 J/s takes
64422540/2205 = 29216 seconds which is
29216/3600 = 8.16 hours
2205 KW * 8.16 hours = 17993 KWh and
17993 * .08 = $1439.44 which seems awfully large, but electric water heaters aren't terribly efficient. ☺
HTH ☺
Doug
2007-02-15 16:08:13 · answer #2 · answered by doug_donaghue 7 · 1⤊ 0⤋