A heater dissipates 1900 W of power when it is connected to a 100 V battery.
What current will flow through the heater when it is connected to the battery?
What is R, the resistance of the heater?
How long does it take to raise the temperature of the air in a good-sized living room that is 6.00 m × 6.00 m × 10.0 m by 10.0∘C? Note that the specific heat of air is 1006 J/(kg⋅∘C) and the density of air is 1.20kg/m3.
Part A -
Current I = Power / Voltage = 1900 / 100 = 19 A
Part B -
Resistance R = V / I = 100 / 19 = 5.26 A
Part C -
Specific heat of dry air = C = 1006 J/(kg C)
Dry air density = p = 1.20 kg/m^3
Room volume = V = 6.00m * 6.00m *10.00m = 360 m^3
Mass of air = m = pV = 360 * 1.20 = 432 kg
Change in temp = delta T = 10C
U = C*m*T
U = 1006 * 432 * 10 = 4345920 J = 4345.92 KJ
Heater capacity = 1900 W
1900 watt = energy / second.
P = U/time ... U must be in J
So, the no. of seconds required to supply this energy, t = 4345920 / 1900 = 2287 s
= 2287/60 = 38.12 min.
Get Answers For Free
Most questions answered within 1 hours.