Calculating intensity and power: How much energy is in a ray of sunlight?

Because power is energy per unit time or $P=\frac{E}{t}$, the definition of intensity can be written as $I=\frac{P}{A}=\frac{E/t}{A}$, and this equation can be solved for E with the given information.

1. Begin with the equation that states the definition of intensity:
   $$I=\frac{P}{A}.$$

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2. Replace $P$ with its equivalent $E/t$:
   $$I=\frac{E/t}{A}.$$

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3. Solve for $E$:
   $$E=\text{IAt}.$$

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4. Substitute known values into the equation:
   $$E=\left(\text{700}{\text{W/m}}^2\right)\left(0\text{.}\text{500}{\text{m}}^2\right)\left[\left(4\text{.}\text{00}\text{h}\right)\left(\text{3600}\text{s/h}\right)\right].$$

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5. Calculate to find $E$ and convert units:
   $$5\text{.}\text{04}\times {\text{10}}^6\text{J},$$

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The energy falling on the solar collector in 4 h in part is enough to be useful—for example, for heating a significant amount of water.

Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.

1. Take the ratio of intensities, which yields:
   $$\frac{I\prime }{I}=\frac{P\prime /A\prime }{P/A}=\frac{A}{A\prime }\left(\text{The powers cancel because}P\prime =P\right)\text{.}$$

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2. Identify the knowns:
   $$A=\text{200}A\prime ,$$

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   $$\frac{I\prime }{I}=\text{200}.$$

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3. Substitute known quantities:
   $$I\prime =\text{200}I=\text{200}\left(\text{700}\mathrm{<mspace\; width="0.25em"><msup><mtext>W/m</mtext><mrow><mn>2</mn></mrow></msup></mspace>}\right).$$

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4. Calculate to find $I\prime$:
   $$I\prime =\text{1.40}\times {\text{10}}^5\mathrm{<mspace\; width="0.25em"><msup><mtext>\; W/m\; </mtext><mrow><mrow><mrow><mn>\; 2\; </mn></mrow></mrow></mrow></msup></mspace>}.$$

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Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.