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Q100PE

Expert-verifiedFound in: Page 1000

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Integrated Concepts**

**(a) On a day when the intensity of sunlight is \({\rm{1}}{\rm{.00\;kW/}}{{\rm{m}}^{\rm{2}}}\), a circular lens \({\rm{0}}{\rm{.200\;m}}\) in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle of \({\rm{20}}{\rm{.}}{{\rm{0}}^{\rm{^\circ }}}\). Assuming the sunlight is unpolarised and the polarizers are \({\rm{100\% }}\)efficient, what is the initial rate of heating of the water in \(^\circ {\rm{C}}/{\rm{s}}\), assuming it is \(80.0\% \) absorbed? The aluminium beaker has a mass of \(30.0\)grams and contains 250 grams of water. **

**(b) Do the polarizing filters get hot? Explain.**

- The initial rate of heating of the water is \(0.02\frac{{^\circ {\rm{C}}}}{{\rm{s}}}\).
- Yes, the polarizing filters get hot.

**Light reflected or transmitted through a particular medium so that all vibrations are confined to a single plane.**

(a)

Let us first calculate the total energy absorbed in one second ( power) by the water and the beaker. Two polarizing sheets of plastic reduce the intensity of the light and then the lens focuses the light on the water and that means that only the light that passes through the lens heats the water and the beaker. The power is then:

\(\begin{align}P &= IA\eta \\ &= {I_0}{\cos ^2}\theta {r^2}\pi \eta \\ &= {10^3}\;\frac{{\rm{W}}}{{{{\rm{m}}^2}}} \times {\cos ^2}{20^\circ } \times {(0.1\;\;{\rm{m}})^2} \times \pi \times (0.8)\\ &= 22.19\;\;{\rm{W}}\end{align}\)

Next, let us find the change in temperature of the water and the beaker in one second. We will assume that the water and the beaker stay in thermal equilibrium at all time (that they have the same temperature).

The total heat that we have to deliver to the water and the beaker is:

\(Q = {m_w}{c_w}\Delta T + {m_a}{c_a}\Delta T\)

If we look at the heat delivered to the water and the beaker in one second we get:

\(\begin{align}\frac{Q}{{\Delta t}} &= {m_w}{c_w}\frac{{\Delta T}}{{\Delta t}} + {m_a}{c_a}\frac{{\Delta T}}{{\Delta t}}\\\frac{Q}{{\Delta t}} &= \frac{{\Delta T}}{{\Delta t}}\left( {{m_w}{c_w} + {m_a}{c_a}} \right)\end{align}\)

But this is just our power delivered \(P\):

\(\begin{align}P &= \frac{{\Delta T}}{{\Delta t}}\left( {{m_w}{c_w} + {m_a}{c_a}} \right)\\\frac{{\Delta T}}{{\Delta t}} &= \frac{P}{{{m_w}{c_w} + {m_a}{c_a}}}\end{align}\)

Plugging in the values (Table 14.1 from page 523. of the textbook) we get:

\(\begin{align}\frac{{\Delta T}}{{\Delta t}} &= \frac{{22.19\;\;{\rm{W}}}}{{(0.25\;\;{\rm{kg}})\left( {4186\;\frac{{\rm{J}}}{{{\rm{kg}}{ \cdot ^\circ }{\rm{C}}}}} \right) + (0.03\;\;{\rm{kg}})\left( {900\;\frac{{\rm{J}}}{{{\rm{kg}}{ \cdot ^\circ }{\rm{C}}}}} \right)}}\\ &= 0.02\;\frac{{^\circ {\rm{C}}}}{{\rm{s}}}\end{align}\)

b) The polarizing filter gets hot. They absorb some of the light energy as molecular kinetic energy (vibration), but temperature is a measure of kinetic energy.

\(\left( {{E_k} = \frac{3}{2}{k_b}T} \right)\)

So, if the kinetic energy increases then the temperature increases as well.

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