Question

i understand the ir sensor will pointed at the ground only, disciplinary action will be taken initials, ____

surface material	temps in direct sun (°c)	average in sun	temps under tree/shade (°c)	average in shade	temp difference (δt)
green grass/plant	30, 27, 36		17, 18, 18
soil or sand	34, 31, 33		14, 9, 12
concrete	30, 30, 26		13, 12, 14
blue glass in quad	29, 29, 29		9, 10, 9

plants open tiny pores on their leaves called stomata to take in co₂. while these pores are water to water vapor absorbs

Explanation:

To solve for the average temperature in sun, average in shade, and temperature difference for each surface material, we follow these steps:

Asphalt (blacktop)

Step 1: Average in Sun

The temperatures in direct sun are 33, 31, 33.
\[
\text{Average in Sun} = \frac{33 + 31 + 33}{3} = \frac{97}{3} \approx 32.33^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 13, 12, 15.
\[
\text{Average in Shade} = \frac{13 + 12 + 15}{3} = \frac{40}{3} \approx 13.33^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 32.33 - 13.33 = 19^\circ\text{C}
\]

Green Grass/Plant

Step 1: Average in Sun

The temperatures in direct sun are 30, 27, 36.
\[
\text{Average in Sun} = \frac{30 + 27 + 36}{3} = \frac{93}{3} = 31^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 17, 18, 18.
\[
\text{Average in Shade} = \frac{17 + 18 + 18}{3} = \frac{53}{3} \approx 17.67^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 31 - 17.67 = 13.33^\circ\text{C}
\]

Soil or Sand

Step 1: Average in Sun

The temperatures in direct sun are 34, 31, 33.
\[
\text{Average in Sun} = \frac{34 + 31 + 33}{3} = \frac{98}{3} \approx 32.67^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 14, 9, 12.
\[
\text{Average in Shade} = \frac{14 + 9 + 12}{3} = \frac{35}{3} \approx 11.67^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 32.67 - 11.67 = 21^\circ\text{C}
\]

Concrete

Step 1: Average in Sun

The temperatures in direct sun are 30, 30, 26.
\[
\text{Average in Sun} = \frac{30 + 30 + 26}{3} = \frac{86}{3} \approx 28.67^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 13, 12, 14.
\[
\text{Average in Shade} = \frac{13 + 12 + 14}{3} = \frac{39}{3} = 13^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 28.67 - 13 = 15.67^\circ\text{C}
\]

Blue Glass in quad

Step 1: Average in Sun

The temperatures in direct sun are 29, 29, 19.
\[
\text{Average in Sun} = \frac{29 + 29 + 19}{3} = \frac{77}{3} \approx 25.67^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 9, 10, 9.
\[
\text{Average in Shade} = \frac{9 + 10 + 9}{3} = \frac{28}{3} \approx 9.33^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 25.67 - 9.33 = 16.34^\circ\text{C}
\]

Final Table

Surface Material	Temps in Direct Sun (°C)	Average in Sun (°C)	Temps under Tree/Shade (°C)	Average in Shade (°C)	Temp Difference (Δt) (°C)
Green Grass/Plant	30, 27, 36	31	17, 18, 18	≈ 17.67	≈ 13.33
Soil or Sand	34, 31, 33	≈ 32.67	14, 9, 12	≈ 11.67	21
Concrete	30, 30, 26	≈ 28.67	13, 12, 14	13	≈ 15.67
Blue Glass in quad	29, 29, 19	≈ 25.67	9, 10, 9	≈ 9.33	≈ 16.34

Answer:

To solve for the average temperature in sun, average in shade, and temperature difference for each surface material, we follow these steps:

Asphalt (blacktop)

Step 1: Average in Sun

The temperatures in direct sun are 33, 31, 33.
\[
\text{Average in Sun} = \frac{33 + 31 + 33}{3} = \frac{97}{3} \approx 32.33^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 13, 12, 15.
\[
\text{Average in Shade} = \frac{13 + 12 + 15}{3} = \frac{40}{3} \approx 13.33^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 32.33 - 13.33 = 19^\circ\text{C}
\]

Green Grass/Plant

Step 1: Average in Sun

The temperatures in direct sun are 30, 27, 36.
\[
\text{Average in Sun} = \frac{30 + 27 + 36}{3} = \frac{93}{3} = 31^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 17, 18, 18.
\[
\text{Average in Shade} = \frac{17 + 18 + 18}{3} = \frac{53}{3} \approx 17.67^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 31 - 17.67 = 13.33^\circ\text{C}
\]

Soil or Sand

Step 1: Average in Sun

The temperatures in direct sun are 34, 31, 33.
\[
\text{Average in Sun} = \frac{34 + 31 + 33}{3} = \frac{98}{3} \approx 32.67^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 14, 9, 12.
\[
\text{Average in Shade} = \frac{14 + 9 + 12}{3} = \frac{35}{3} \approx 11.67^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 32.67 - 11.67 = 21^\circ\text{C}
\]

Concrete

Step 1: Average in Sun

The temperatures in direct sun are 30, 30, 26.
\[
\text{Average in Sun} = \frac{30 + 30 + 26}{3} = \frac{86}{3} \approx 28.67^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 13, 12, 14.
\[
\text{Average in Shade} = \frac{13 + 12 + 14}{3} = \frac{39}{3} = 13^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 28.67 - 13 = 15.67^\circ\text{C}
\]

Blue Glass in quad

Step 1: Average in Sun

The temperatures in direct sun are 29, 29, 19.
\[
\text{Average in Sun} = \frac{29 + 29 + 19}{3} = \frac{77}{3} \approx 25.67^\circ\text{C}
\]

Step 2: Average in Shade

The temperatures under tree/shade are 9, 10, 9.
\[
\text{Average in Shade} = \frac{9 + 10 + 9}{3} = \frac{28}{3} \approx 9.33^\circ\text{C}
\]

Step 3: Temperature Difference

\[
\text{Temperature Difference} = 25.67 - 9.33 = 16.34^\circ\text{C}
\]

Final Table

Surface Material	Temps in Direct Sun (°C)	Average in Sun (°C)	Temps under Tree/Shade (°C)	Average in Shade (°C)	Temp Difference (Δt) (°C)
Green Grass/Plant	30, 27, 36	31	17, 18, 18	≈ 17.67	≈ 13.33
Soil or Sand	34, 31, 33	≈ 32.67	14, 9, 12	≈ 11.67	21
Concrete	30, 30, 26	≈ 28.67	13, 12, 14	13	≈ 15.67
Blue Glass in quad	29, 29, 19	≈ 25.67	9, 10, 9	≈ 9.33	≈ 16.34