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Chapter 6: Q87GP (page 138)

Electric energy units are often expressed in 鈥渒ilowatt-hours.鈥� (a) Show that one kilowatt-hour (kWh) is equal to\(3.6 \times {10^6}\;{\rm{J}}\). (b) If a typical family of four uses electric energy at an average rate of 580 W, how many kWh would their electric bill show for one month, and (c) how many joules would this be? (d) At a cost of $0.12 per kWh, what would their monthly bill be in dollars? Does the monthly bill depend on the rate at which they use the electric energy?

Short Answer

Expert verified

(b) 420 kWh (c)\(1.5 \times {10^9}\;{\rm{J}}\)(d) $50

No, the monthly bill does not rely upon the rate at which they use electrical energy.

Step by step solution

01

Understand the relation betweenpower and work

In this problem, the standard relation between electrical energy and power with time will be helpful in estimating the required values either in Joule or kilo-watt-hour.

Power (P) is defined as the rate of doing work (W) or consuming energy (E).

\(P = \frac{W}{t} = \frac{E}{t}\)

Given data:

The average rate of electrical energy is\(P = 580\;{\rm{W}}\).

The cost per kWh is .

02

Part (a) convert energy intoJoules

The relation to calculate the energy in Joulesis given by:

\(\begin{aligned}1\;{\rm{kWh}} &= \left( {1\;{\rm{kWh}} \times \frac{{1000\;{\rm{W}}}}{{1\;{\rm{kW}}}} \times \frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}} \times \frac{{1\;{\rm{J/s}}}}{{1\;{\rm{W}}}}} \right)\\1\;{\rm{kWh}} &= 3.6 \times {10^6}\;{\rm{J}}\end{aligned}\)

Thus, it is proved that \(1\;{\rm{kWh}} = 3.6 \times {10^6}\;{\rm{J}}\).

03

Part (b) calculation of the energy for one month in kWh

The relation to calculate the energy (E) is given by:

\(E = P \times t\)

Here,\(t\)is the time whose value is onemonth.

On plugging the values in the above relation, you get:

\(\begin{aligned}E &= \left( {580\;{\rm{W}} \times \frac{{1\;{\rm{kW}}}}{{1000\;{\rm{W}}}}} \right) \times \left( {1\;{\rm{month}} \times \frac{{30\;{\rm{day}}}}{{1\;{\rm{month}}}} \times \frac{{24\;{\rm{h}}}}{{1\;{\rm{day}}}}} \right)\\E &= 417.6\;{\rm{kWh}}\\E &\approx 420{\rm{ kWh}}\end{aligned}\)

Thus, \(E = 417.6\;{\rm{kWh}}\) is the required energy.

04

Part (c) calculation of the energy in Joules

The relation to calculate the energy in Joules is given by:

\(\begin{aligned}E &= 417.6\;{\rm{kWh}} = \left( {417.6\;{\rm{kWh}} \times \frac{{1000\;{\rm{W}}}}{{1\;{\rm{kW}}}} \times \frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}} \times \frac{{1\;{\rm{J/s}}}}{{1\;{\rm{W}}}}} \right)\\E &= 417.6\;{\rm{kWh}} = 1.5 \times {10^9}\;{\rm{J}}\end{aligned}\)

Thus, \(417.6\;{\rm{kWh}} = 1.5 \times {10^9}\;{\rm{J}}\)is the required energy in kWh.

05

Calculate the monthly bill in dollars

The relation to calculate the monthly bill (\({P_{\rm{B}}}\))is given by:

\({P_{\rm{B}}} = E \times c\)

Here,cis the cost per unit or cost per kWh

On plugging the values in the above relation, you get:

\(\begin{aligned}{P_{\rm{B}}} &= \left( {417.6\;{\rm{kWh}}} \right) \times \left( {\$ 0.12{\rm{/kWh}}} \right)\\{P_{\rm{B}}} &= \$ 50.112\\{P_{\rm{B}}} &= \$ 50\end{aligned}\)

Thus,\({P_{\rm{B}}} = \$ 50\)is the monthly bill in dollars.

The monthly bill does not rely on the rate at which the family consumes electrical energy because kilowatt-hours is not a unit of power or rate.

Therefore, the rate at which energy is consumed does not appear in the monthly bill.

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