Sunlight is not equally intense at all wavelengths, but its Intensity distributi

Question

Sunlight is not equally intense at all wavelengths, but its Intensity distribution is as illustrated in the
accompanying Excel file. Use this data to determine the rate at which thermal energy is transferred to
an object when the sunlight is filtered to only allow
(a) wavelengths between 400 and 450 nm
(b) wavelengths between 500 and 550 nm.
(c) wavelengths between 650 and 750 nm.
(d) wavelengths over the entire visible spectrum, 400 to 750 nm.
(e) If an object of area 1.00 m 2 is exposed to the filtered light for each of the four wavelength ranges
above, calculate the exposure time needed to absorb 1000. J.

(nm)I(W/m2) 395.0 396.0 397.0 398.0 399.0 400.0 401.0 0.81 0.76 0.43 0.85 1.07 1.11 1.16 Sunlight Intensity at Earth's Surface 10 1.80 402.0 1,21 12 1.60 403.0 404.0 405.0 406.0 407.0 408.0 409.0 410.0 411.0 412.0 413.0 414.0 415.0 416.0 417.0 418.0 419.0 420.0 421.0 1.16 1.18 1.15 1.12 1.10 1.15 1.23 1.05 1.17 1.25 1.20 1.18 1.23 1.26 1.23 1.18 1.23 1.12 1.28 1.40 16 17 18 19 1.20 E 1.00 0.80 21 0.60 23 0.40 25 27 0.20 29 30 31 0.00 390.0 440.0 490.0 540.0 590.0 640.0 690.0 740.0 790.0 Wavelength (nm)

Explanation / Answer

Matlab code to plot all three graphs for first three questions is as follows;

wavelength1 = 400:1:450 ;
I1 = [1.11 1.16 1.21 1.16 1.18 1.15 1.12 1.10 1.15 1.23 1.05 1.17 1.25 1.20 1.18 1.23 1.26 1.23 1.18 1.23 1.12 1.28 1.26 1.22 1.21 1.25 1.21 1.17 1.18 1.10 0.87 0.79 1.32 1.23 1.14 1.25 1.37 1.39 1.22 1.18 1.35 1.33 1.43 1.45 1.41 1.46 1.31 1.49 1.51 1.50 1.56];
plot(wavelength1,I1);
Int1 = trapz(wavelength1,I1);

wavelength2 = 500:1:550;
I2 = [1.55 1.50 1.50 1.57 1.46 1.56 1.63 1.56 1.52 1.59 1.55 1.58 1.62 1.52 1.49 1.53 1.55 1.26 1.44 1.40 1.52 1.53 1.57 1.48 1.59 1.58 1.53 1.34 1.54 1.61 1.54 1.63 1.60 1.43 1.53 1.55 1.62 1.50 1.57 1.54 1.48 1.43 1.55 1.53 1.58 1.54 1.53 1.55 1.50 1.55 1.54];
plot(wavelength2,I2);
Int2 = trapz(wavelength2,I2);

wavelength3 = 650:1:750;
I3 = [1.36 1.44 1.39 1.43 1.42 1.35 1.19 1.24 1.39 1.39 1.40 1.39 1.38 1.38 1.40 1.42 1.42 1.41 1.42 1.44 1.42 1.42 1.40 1.41 1.41 1.40 1.41 1.40 1.41 1.39 1.40 1.39 1.40 1.38 1.37 1.37 1.34 0.97 1.12 1.13 1.18 1.23 1.27 1.26 1.25 1.27 1.27 1.34 1.32 1.29 1.28 1.27 1.27 1.27 1.31 1.32 1.31 1.31 1.30 1.31 1.32 1.32 1.31 1.29 1.30 1.26 1.27 1.11 1.03 0.92 0.99 0.09 0.24 1.14 1.06 1.04 1.08 1.09 1.04 1.05 1.13 1.07 1.15 1.20 1.24 1.22 1.21 1.20 1.23 1.19 1.22 1.21 1.22 1.24 1.25 1.25 1.25 1.25 1.24 1.24 1.23];
plot(wavelength3,I3);
Int3 = trapz(wavelength3,I3);

I have calculated area under the curve through trapezoidal method.

Thermal energy transferred is the area under the curve so, following are the answers for respective questions.

(a) 61.7750 W/m2 which is Int1 in above code.

(b) 76.4150 W/m2 which is Int2 in above code.

(c) 125.325 W/m2 which is Int3 in above code.

(d) add the total area for above parts which is 263.515 W/m2

(e) exposure time needed to absorb 1000 J for part (a) = 1000/61.7750 = 16.1877 seconds

      exposure time needed to absorb 1000 J for part (b) = 1000/76.4150 = 13.0864 seconds

      exposure time needed to absorb 1000 J for part (c) = 1000/125.325 = 7.9792 seconds

      exposure time needed to absorb 1000 J for part (e) = 1000/263.515 = 3.7948 seconds

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