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3 years ago

How does a greenhouse keep radiant energy from escaping?

1 answer:

7 0

Greenhouse effect is the process of trapping sun radiation in the earth surface, so as to make the planet warm. It is a natural phenomenon in which sun rays that enters the earth surface are re-radiated into the earth surface causing the heat trap in the earth.
In similar manner, green house also trap radiation inside the glass house. The sun rays once reach inside the glass house, strike on plants and objects and reflected back. For entire radiation to escape from the green house, the heat flow between the two points must be proportional to the temperature difference and thermal conductivity. The glass of which the green house are made up, have low thermal conductivity. So, temperature must rise inside glass house so to maintain the heat flow rate of incoming and outgoing radiation.
The green house glass are insulated and they trap the infrared emitted by the objects inside the green house from escaping outside. Since the infrared have longer wave lengths, it is released slowly.

Also, the thermal energy inside the glass house are transferred by convection process. But the glass walls and roof act as insulator, keeping the radiant energy from escaping outside the green house.<span />

You might be interested in

What is the area of triangle ABC with vertices A(x¹,y¹), B(x²,y²)and C (x³,y³)??????????<br>

AlexFokin [52]

Answer:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

$Area = 2\ units^2$

Explanation:

Given

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

Required

Determine the area

The area of a triangle is :

$Area = \frac{1}{2}|A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|$

By substituting values for the x and y coordinates of A, B and C;

We have:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

So:

For instance

$A = (0,3)$

$B= (2,1)$

$C = (2,3)$

The area is:

$Area = \frac{1}{2}|0(1-3) + 2(3-3) + 2(3-1)|$

$Area = \frac{1}{2}| 2*0 + 2*2|$

$Area = \frac{1}{2}| 0 + 4|$

$Area = \frac{1}{2}|4|$

$Area = \frac{1}{2} * 4$

$Area = 2\ units^2$

7 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$