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(I) Solar cells (Fig. 15–26) can produce about 40 W of electricity per square meter of surface area if directly facing the Sun. How large an area is required to supply the needs of a house that requires 24 kWh/day? Would this fit on the roof of an average house? (Assume the Sun shines about 9 h/day.).

FIGURE 15-26 Problem 53

Short Answer

Expert verified

The area required to supply the needs of the house is \[66.67\;{{\rm{m}}^2}\], and the solar cells would fit on the roof of an average house.

Step by step solution

01

Understanding solar cells

Solar cells are devices used to directly convert light energy (sunlight) into electricity. They do not require any heat engine.

02

Given data

The power per meter square is \(P = 40\;{\rm{W/}}{{\rm{m}}^2}\).

The supply given to the house is \(E = 24\;{\rm{kWh/day}}\).

The sun shines about \(n = 9\;{\rm{h/day}}\).

03

Evaluation of the area required to supply the need of the house

The area required to supply the need of the house is calculated as:

\(\begin{array}{l}A = \left( {24\;{\rm{kWh/day}}} \right)\left( {\frac{{1\;{\rm{day}}}}{{9\;{\rm{h}}}}} \right)\left( {\;\frac{{{\rm{1}}{{\rm{m}}^2}}}{{40\;{\rm{W}}}}} \right)\\A = \left( {24 \times {{10}^3}\;{\rm{Wh/day}}} \right)\left( {\frac{{1\;{\rm{day}}}}{{9\;{\rm{h}}}}} \right)\left( {\;\frac{{{\rm{1}}{{\rm{m}}^2}}}{{40\;{\rm{W}}}}} \right)\\A = 66.67\;{{\rm{m}}^2}\end{array}\)

Thus, the area required to supply the need of the house is \(66.67\;{{\rm{m}}^2}\).

04

Evaluation of the roof area of the house

terrace at a \(30^\circ \) angle.

The area of the roof is calculated as:

\(\begin{array}{l}{A_{\rm{R}}} = \left( {1000\;{\rm{f}}{{\rm{t}}^2}} \right)\left( {\frac{1}{{\cos 30^\circ }}} \right){\left( {\;\frac{{{\rm{1m}}}}{{3.28\;{\rm{ft}}}}} \right)^2}\\{A_{\rm{R}}} = 107.33\;{{\rm{m}}^2}\end{array}\)

The area of the roof for a small house is \({A_{\rm{R}}} = 107.33\;{{\rm{m}}^2}\), which is larger than the required area of the house that is \(A = 66.67\;{{\rm{m}}^2}\). Thus, the set of cells will fit on the roof of the house.

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Most popular questions from this chapter

In an isothermal process, 3700 J of work is done by an ideal gas. Is this enough information to tell how much heat has been added to the system? If so, how much? If not, why not?

Which of the following possibilities could increase the efficiency of a heat engine or an internal combustion engine?

(a)Increase the temperature of the hot part of the system and reduce the temperature of the exhaust.

(b) Increase the temperatures of both the hot part and the exhaust part of the system by the same amount.

(c) Decrease the temperatures of both the hot part and the exhaust part of the system by the same amount.

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(e) None of the above; only redesigning the engine or using better gas could improve the engine's efficiency.

Question:(II) A heat engine exhausts its heat at 340°C and has a Carnot efficiency of 36%. What exhaust temperature would enable it to achieve a Carnot efficiency of 42%?

Question: An ideal gas undergoes an isothermal expansion from state A to state B. In this process (use sign conventions, page 413),

(a) \[Q = 0,\;\Delta U = 0,\;W > 0\].

(b) \[Q > 0,\;\Delta U = 0,\;W < 0\].

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(d) \[Q > 0,\;\Delta U = 0,W > 0\].

(e) \[Q = 0,\;\Delta U < 0,\;W < 0\].

Can mechanical energy ever be transformed completely into heat or internal energy? Can the reverse happen? In each case, if your answer is no, explain why not, if yes, give one or two examples.

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