A heating coil has a resistance of 4.30 ohms and operates on 120 V. The current in the coil while it is operating is 27.91 amps.
What energy is supplied to the coil in 13 min?
If the coil is immersed in an insulated container holding 27.0 kg of water, what will be the increase in the temperature of the water? Assume that 100% of the heat is absorbed by the water. The specific heat of water is 4180 J/(kg*C°).
At $0.09 per kWh, how much does it cost to operate the heating coil 20 min per day for 30 days (in dollars)?
Can anyone help me as to how to solve this problem??
2007-04-04 12:45:50 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
RayM is right but I hope you would like to know how and why.
Power = E*I = 120V*27.91a = 3.349 kW
Energy can be measured in several different units. One of them is kW*hours. Your electric bill comes with those units. So convert 13 minutes to hours and multiply.
2007-04-04 13:15:54 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
Resistance, R = 4.3 ohms
Volt, V = 120 V
Current, I = 27.91
Energy supplied to the coil in 13 min -
t = 13 min = 780 sec
Work done,
W = V * I * t = 2612376 J = 2612.38 kJ
Energy supplied = Work done = 2612.38 kJ
When the coil is immersed in water -
Heat gained by water = Heat lost by the coil = Energy possessed by the coil
Heat gained by water, Q = 2612.38 kJ
Mass of water, m = 27 Kg
Specific Heat Capacity of water, C = 4.18 kJ
Rise in temperature of water, dt = ?
Q = m * C * dt
dt = 23.15 degrees
Monthly cost to operate -
Power of the coil, P = V * I = 3349.2 W = 3.349 kW
Time for which it is used, t = 20 * 30 min = 10 hr
Energy consumed, E = P * t = 33.49 kWh
Cost for
1 kWh = $0.09
33.49 kWh = $3.014
2007-04-06 08:02:35 · answer #2 · answered by Prashant 6 · 0⤊ 0⤋
13/60*120V*27.91=___KWH
I want to buy at $.09/KWH
2007-04-04 19:55:15 · answer #3 · answered by RayM 4 · 0⤊ 0⤋