Question

# A heater dissipates 1900 W  of power when it is connected to a 100 V  battery. Part A...

A heater dissipates 1900 W  of power when it is connected to a 100 V  battery.

Part A

What current will flow through the heater when it is connected to the battery?

Part B

What is R, the resistance of the heater?

Part C

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.

#### Homework Answers

Answer #1

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.

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