Question

12.5 x 10J c) How long will it take to heat a room from 10 Cto2Cif the heater consumes d) How many kilowatt-hours of electricity are used during the time calculated in part (c)? e) What is the cost (with electrical energy at 25 cents per kWh) of running the heater? 6. A metal wire has a resistance ofB1 Qunder coom temperature conditions of20°c)When the wire is he esistance increases by0.70 )What is the temperature coefficient of resistivity of this metal? 45x1033 copper cable needs to carry a cument.of170 A with a power loss of only 6.0W/m, What is the reg er cable? (The resistivity of copper is 1.7 x 10-8 Qm)

Answer 1

How long will it take to heat a room from 10 ⁰C to 20 ⁰C if the heater consumes 12.5 X 10⁵ J?

Specific heat capacity of air is 1.006 kJ / kg ⁰C

To calculate mass of air:
     density of air is 0.036 kg / ft³

Let the total volume of the room is 10 x 10 x 10 = 1000 ft³

Thus mass = density x volume

= 0.037 kg / ft³x 1000  ft³ = 37 kg

Temperature difference = 10 ⁰C (20 ⁰C - 10 ⁰C)

Energy needed, Q = mass x specific heat capacity of air x change in temperature

Q = 37 kg x 1.006 kJ / kg ⁰C x 10 ⁰C

= 372.22 kJ

Power is energy transferred / second.

Assuming, the heater consumes 12.5 x 10⁵ J of heat in one hour, then time required to heat the room is calculated as follows.

12.5 x 10⁵ J --> 3600 second

372.22 kJ ---> a

$a = \frac{372.22 \times 10^{3} \times 3600}{12.5\times 10^{5}}$

thus a = 0.2977 hour.

d. How many kilowatt hours of electricity is used during 0.2977 hour?

$Power = \frac{Energy}{Time} \\ \\ Power = \frac{272.22 \ kJ}{1 \ s} \\ \\ Power = 272.22 \ \ W$

Thus Electrical energy = power x time

= 0.2722 kW x 0.2977 hour

= 0.081 kWh

With 25 cents per kWh, the cost of running the heater is = 0.081 kWh x 25 cents / kWh

= 2.025 cents