2.2 Spherical mirrors  (Page 6/20)

1. If we want the rays from the sun to focus at 40.0 cm from the mirror, what is the radius of the mirror?
2. What is the amount of sunlight concentrated onto the pipe, per meter of pipe length, assuming the insolation (incident solar radiation) is 900 ${\text{W/m}}^2$ ?
3. If the fluid-carrying pipe has a 2.00-cm diameter, what is the temperature increase of the fluid per meter of pipe over a period of 1 minute? Assume that all solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.

Strategy

Solution

1. The sun is the object, so the object distance is essentially infinity: $d_{\text{o}}=\text{∞}$ . The desired image distance is $d_{\text{i}}=40.0\text{cm}$ . We use the mirror equation to find the focal length of the mirror:
   
   $\begin{array}{cc}\hfill \frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}& =\frac{1}{f}\hfill \\ \hfill f& ={\left(\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}\right)}^{\mathrm{-1}}\hfill \\ & ={\left(\frac{1}{\text{∞}}+\frac{1}{40.0\text{cm}}\right)}^{\mathrm{-1}}\hfill \\ & =40.0\text{cm}\hfill \end{array}$
   
   Thus, the radius of the mirror is $R=2f=80.0\text{cm}$ .
2. The insolation is 900 ${\text{W/m}}^2$ . You must find the cross-sectional area A of the concave mirror, since the power delivered is 900 ${\text{W/m}}^2\times A$ . The mirror in this case is a quarter-section of a cylinder, so the area for a length L of the mirror is $A=\frac{1}{4}(2\pi R)L$ . The area for a length of 1.00 m is then
   
   $A=\frac{\pi }{2}R(1.00\text{m})=\frac{(3.14)}{2}(0.800\text{m})(1.00\text{m})=1.26{\text{m}}^2.$
   
   The insolation on the 1.00-m length of pipe is then
   
   $\left(9.00\times {10}^2\frac{\text{W}}{{\text{m}}^2}\right)\left(1.26{\text{m}}^2\right)=1130\text{W}.$
3. The increase in temperature is given by $Q=mc\text{Δ}T$ . The mass m of the mineral oil in the one-meter section of pipe is
   
   $\begin{array}{cc}\hfill m& =\rho V=\rho \pi {\left(\frac{d}{2}\right)}^2(1.00\text{m})\hfill \\ & =\left(8.00\times {10}^2{\text{kg/m}}^3\right)(3.14){(0.0100\text{m})}^2(1.00\text{m})\hfill \\ & =0.251\text{kg}\hfill \end{array}$
   
   Therefore, the increase in temperature in one minute is
   
   $\begin{array}{cc}\hfill \text{Δ}T& =Q\text{/}mc\hfill \\ & =\frac{\left(1130\text{W}\right)\left(60.0\text{s}\right)}{\left(0.251\text{kg}\right)\left(1670\text{J}·\text{kg/°C}\right)}\hfill \\ & =162\text{°C}\hfill \end{array}$

Significance