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Chapter 2: Q74E (page 67)

A typical household uses 500 kWh of energy in I month. How much mass is convened to produce this energy?

Short Answer

Expert verified

The mass to produce energy for household consumption is $2 \times 10^{- 8} kg$ .

Step by step solution

01

Identification of given data

The energy consumption of household is $E = 500 kWh$

02

Energy Consumption

The energy used by all the electrical appliances of house hold in a month for working hour is electrical consumption. This energy is equated to rest energy of mass used.

03

Determination of mass to produce energy for household consumption

The mass to produce energy for household consumption is given as:

$E = Δ m c^{2} Δ m = \frac{E}{c^{2}}$

Here, c is the speed of light and its value is $3 \times 10^{8} m / s$ ,

Substitute all the values in the equation.

$Δ m = \frac{(500 kWh) (\frac{10^{3} J / s}{1 kW}) (\frac{3600 s}{1 h})}{{(3 \times 10^{8} m / s)}^{2}} Δ m = 2 \times 10^{- 8} kg$

Therefore, the mass to produce energy for household consumption is $2 \times 10^{- 8} kg$ .

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Most popular questions from this chapter

From the Lorentz transformation equations, show that if time intervals between two events, $∆ t$ and $∆ t'$ , in two frames are of opposite sign, then the events are too far apart in either frame for light to travel from one to the other. Argue that therefore they cannot be casually related.

Prove that if v and u' are less than c, it is impossible for a speed u greater than c to result from equation (2-l9b). [Hint: The product (c-u')(c-v) is positive.]

Explain to your friend, who is willing to accept that light moves at the same speed in any frame, why clocks on a passing train are not synchronized. If it helps, assume that Anna is at the middle of the train.

Derive the following expressions for the components of acceleration of an object , $a_{x}'$ and $a_{y}'$ in the frame $S'$ in terms of its components of acceleration and velocity in the frame $S$ .

${a_{x}}^{'} = \frac{a_{x}}{{γ_{v}}^{3} {(1 - \frac{u_{x} v}{c^{2}})}^{3}} {a_{y}}^{'} = \frac{a_{y}}{{γ_{v}}^{2} {(1 - \frac{u_{x} v}{c^{2}})}^{2}} + \frac{a_{x} \frac{u_{y} v}{c^{2}}}{{γ_{v}}^{2} {(1 - \frac{u_{x} v}{c^{2}})}^{3}}$

Show that for a source moving towards an observer equation (2-17) becomes $f_{o b s} = f_{s o u r c e} \sqrt{\frac{1 + v / c}{1 - v / c}}$

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