Unit One Help

Science 123		Spring 2020
Unit One Homework Help

Chapter One - Atmospheric Variables

6.	Use the Ideal Gas Law. Remember, the temperature $T$ must be in Kelvins.
8.	(b) The elevation must be converted to meters, with 1 mile = 1610 m.
9.	The formula for potential temperature $\theta$ is $\displaystyle{\theta = T\left(\frac{1000}{P} \right)^{0.2859}}$. The parcel temperature $T$ must be in Kelvins. Here the $T=-5+273 = 268$. Pressure $P$ is in mb, so $P$=700. Leave your answer in Kelvins. Answer: 296.8 K.
10.	Use the relationship between parcel temperature $T$ and potential temperature $\theta$ to determine the pressure of the parcel. Remember, both $T$ and $\theta$ must be in Kelvin. In the process of solving $\displaystyle{\theta = T\left(\frac{1000}{P} \right)^{0.2859}}$ for $P$. an intermediate step would be $\displaystyle{\left(\frac{\theta}{T} \right)^{3.498} = \frac{1000}{P}}.$ (NOTE: 1/0.2859 = 3.498). After subbing for $T$ and $\theta$, you should get $P$ value of around 904 mb. Once the pressure is known, determine the height through the height and pressure relationship: $\displaystyle{z = -8000\ln\left(\frac{P}{1000}\right)}$
12.	Use the formula for density as a function of height : $\rho(z) = 1.22e^{-z/8000}$ and solve for $z$ by dividing both sides of the equation by 1.22 and then taking the natural log of both side of the resulting equation. That is, $\ln( \rho/1.22) = -z/8000$ or $z = -8000 \ln(\rho/1.22)$.
14.	Remember (i) that the percent by volume and the present by mass for oxygen (or any constituent) are approximately the same, and (ii) if not specified, assume a surface density of 1.22 g/L or kg/$\mbox{m}^3$. The volume of the room is 360 m$^3$. Multiply this value time the density of 1.22 kg per m$^3$ to get the mass of air.
16.	Calculate $\frac{\Delta T}{\Delta z}$ to find the lapse rate, an report it in change of temperature per 1000 meter. Remember, the lapse rate is a positive number if temperature decreases with height.
18.	The surface temperature is $\displaystyle{\frac{5}{9}(72 - 32) = 23.9}$ C $\displaystyle{T(3000) = 23.9 - \frac{4}{1000} \times 1500 - \frac{1}{1000} \times 500 - \frac{3}{1000} \times 1000 = 14.4}$ C. follow a similar approach to determine the temperature at the other elevations.
19.	Use the Ideal Gas Law to determine the surface pressure first. Then use $\displaystyle{z = -8000\ln\left(\frac{1000}{P}\right)}$ to determine the height. The height over Decorah is approximately 5692 m, while the elevation of the 500 mb level over Tallahassee is approximately 5750 m. Note that the height is higher at the region where the surface temperature is higher.
20.	Use the Ideal Gas Law to determine the surface pressure first. The use $\displaystyle{z = -8000\ln\left(\frac{1000}{P}\right)}$ to determine the height of the 600 mb level.
22.	The column between 1000 mb and 500 mb can be broken into three segments : 1000 mb - 850 mb, 850 mb - 700 mb, 700 mb - 500 mb. The mixing ratio for a given segment is best given by the average of the top and bottom values of the given segment. For example, use a mixing ratio of (20+15)/2 = 17.5 gm/kg for the first segment. To determine the mass of water vapor in the first segment, the following short cut can be used. The total mass of dry air in the column is approximately 10,200 kg. At the 850 mb level, 15% of the total mass is below that height because (1000 - 850)/1000 = 0.15, or 15% of the columnar mass. So 0.15$\times$10,200 = 1530 kg. The mass of water vapor in this segment of the column is 17.5 $\times$ 1530 = 26,775 gm or 26,775 cc. Follow a similar method for the remaining two segments to determine the volume of water vapor in each. The final answer is 6.12 cm.
23.	(a) The mass of water vapor is 32,940 grams. (b) The volume of water is 32,940 cc. The surface area of the room is 300 $\mbox{m}^2$ = 300 $\times$ 100 $\times$ 100 = 3,000,000 $\mbox{cm}^2$ because there are 100 cm in a meter. The depth of water is 32,940/3,000,000 = 0.011 cm.
24.	Work on the basis of a 1 $\mbox{m}^2$ column of air. The depth of rain in cm is 1.5 $\times$ 2.54 = 3.81 cm. This is the precipitated vapor. The depth if all the vapor precipitated would be twice this depth, or 7.62 cm. The volume (or mass) of vapor in the column is 7.62 $\times$ 100 cm $\times$ 100 cm = 76,200 cc (grams). Next, we need the mass of dry air in the column. The total mass in the column is 10204 kg. The mass above 6000 m is mass = 10204$e^{-6000/8000}$ = 4820 kg. Therefore, the mass between the surface and 6000 m is 10204-4820=5384 kg. The mixing ratio is 76200/5384 = 14.2 gm/kg.

Chapter Two - Energy Forms and Transfer

1.	(a) Remember, 1 watt is a joule per second, so a 100-watt bulb gives of 100 joules each second. (b) The conversion of pounds to kilograms is given on the equation/constants/conversions sheet handed out the first day.
2.	(a) For purposes of PE and KE, the mass must be measured in kilograms. (b) $PE = m \times g \times z$, where $z$ must be in meters. (c) $KE = \frac{1}{2} \times m \times v^2$, where mass $m$ is in kilograms and velocity $v$ is in meters per second. (d) The other energy form is thermal energy, of course. Additionally, there is friction that may dissipate KE.
3.	(a) Use $\Delta Q = mL$, latent heat equation (the change in energy due to phase change of H₂O). Here, $L$ = 600 cal/gm. (b) Use $\Delta Q = mC\Delta T$, where $C$ is for dry air.
5.	(a) One mile = 1610 meters. So 1 square mile = $1610 \times 1610$ square meters. Now, each square meter is $100 \times 100$ square centimeters. (b) You should find the energy released during the thunderstorm is _________ $\times 10^{15}$ J. (c) The latent heat released in this "average" thunderstorm is considerably more (many times more) than that associated with the atomic bomb. In fact, that ratio of energies = $\displaystyle{\frac{Q_{storm}}{Q_{bomb}} \approx 12.5}$
6.	(a) The number of seconds in a day is 60 sec/min $\times$ 60 min/hr $\times$ 24 hr/day = 86,400. You should find the total energy emitted is $3.484 \times 10^7$ joules. (b) You should find that the energy from the one-meter square region would supply approximately 32% of the average daily household use.
8.	net radiation = radiation in - radiation out $\Rightarrow$ $R_{net} = R_{in} - R_{out}$ $\displaystyle{R_{net} = \sigma T_{a}^4 \times \mbox{area} \times \mbox{ time (in seconds)} - \sigma T_{b}^4 \times \mbox{area} \times \mbox{ time (in seconds)} }$. $T_b$ = the body temeprature in K, and $T_a$ is the air temperature in K. Time (in seconds) is the length of a day (24 $\times$ 60 $\times$ 60). You should find a net loss between 2,000 and 3,000 kilocalories.
10.	(a) Note that $\pi$ and $r^2$ cancel out, so they may be removed from both sides of the equation. Substituting values and calculating the result gives $\epsilon$ = 0.5871. (b) The new albedo of 0.30 and the $\epsilon$ from part (a) result in a new surface average temperature of 18.19 C.

Chapter Four: Forces & Winds

1.	(a) 11.18 m /sec (b) (2, -7) (c) $\displaystyle{\vec{a} = \frac{\Delta \vec{v}}{\Delta t} = (2/3600, -7/3600) = (0.00056, -0.00194) }$ m / sec$^2$. (d) $\displaystyle{\vec{F} = m \times \vec{a} = 2 \times (0.00056, -0.00194) = (0.0011, -0.0039)}$ N.
2.	(a) The trough runs diagonally, on the left-hand-side of the figure, from near the top to the bottom. The ridge axis runs pretty much parallel to the trough axis, from the top center portion of the figure to the lower right-hand corner (near the "N") of the figure. (b) Winds at the surface are not geostrophic nor gradient due to the influence of friction. Friction diminishes the speed of the air parcels. As the speed decreases, the coriolis effect decreases, so the right-ward deflection is less (the relative effect of the pressure gradient force is greater). Consequently, the parcels do not follow the contours as they would in geostrophic or gradient flow, but track slightly across the pressure contours toward the lower pressure. Here is a link that illustrates this effect. (c) Remember to convert pressure to N/m². Use the scale on the lower right hand corner of the figure to gauge $\Delta x$. The pressure gradient at point A is approximately $8 \times 10^{-4}$ N/m²/m. (e) The geostrophic wind speed at point A is approximately 6.8 m/sec. (g) Recall the geostrophic wind does not account for friction.
4.	(a) The hydrostatic equation is $\frac{\Delta P}{\Delta z} = -\rho g$. In this case, $\Delta z$ = 5600 m, $\rho$ = 0.91 kg/m³, and $g$ = 9.8 m/sec. The change in pressure is -49941 N/m². Change this to millibars to get the pressure at 5600 meters. Here is a link to a related example using the hydrostatic equation. (b) A colder column of air "collapses" or compresses downward. Therefore, the pressure levels compress as well since pressure can correctly be interpreted as a measure of the mass above a given point in the column. (c) Follow the same method as that in part (a). Answer : Pressure at 5600 m is 467.7 mb.
5.	You can determine the pressure gradient by knowing the geostrophic wind speed. Knowing the pressure gradient and $\Delta x$ will allow you to determine the $\Delta P$ in N/m². Once you have $\Delta P$ in mb, you will need to determine if it should be added or subtracted from the 1000 mb reading at point B. Determine this from the wind direction and the fact that the Coriolis deflection (force) points to the right of the wind direction and the pressure gradient force points in the opposite direction AND from high to low pressure.
6.	The gradient wind equation for flow around a low or through a trough of low pressure should reduce to $\displaystyle{v^2 + 20.6v - 279}$ in simpler form. Using the quadratic equation with the positive option of the $\pm$ term, the gradient wind speed you find should be less than 10 m/sec and greater than 5 m/sec.