Quiz2_practice_2023t1_solutions

EOSC 340 Quiz 2 practice problems 2022t1 1 /12 This quiz will cover all topics from day 9 (climate feedbacks and sensitivity) through day 19 (Modern Carbon cycle), including material covered in assignments 3, 4 and 5. To help you do well on the quiz, work through these practice problems, look over the worksheets and the solutions, and go through the class slides. Think about how similar questions could be made, and try to answer those too. 1. Ice-albedo feedback can be written: ࠵? = # ࠵?࠵? ࠵?ice area , × # ࠵?ice area ࠵?࠵? , Which of the following statements is correct? A) / 01 0ice area 2 is positive, / 0ice area 03 2 is positive, ࠵? is positive B) / 01 0ice area 2 is positive, / 0ice area 03 2 is negative, ࠵? is negative C) / 01 0ice area 2 is negative, / 0ice area 03 2 is positive, ࠵? is negative D) / 01 0ice area 2 is negative, / 0ice area 03 2 is negative, ࠵? is positive E) / 01 0ice area 2 is negative, / 0ice area 03 2 is negative, ࠵? is negative Remember that a positive increase in ice area is cooling, and that ice area decreases with a positive temperature change. 2. If the change in radiative forcing per square kilometer of arctic ice ( ! R / ! (ice area) ) is -10 -6 (W m -2 ) / km 2 , how much ice would have to melt per degree of warming ( ! (ice area)/ ! T ) to make the ice- albedo feedback strength c A = 0.25 W m -2 K -1 ? A) -0.25 x 10 6 km 2 / K B) -0.5 x 10 6 km 2 / K C) 1.5 x 10 6 km 2 / K D) 2.0 x 10 6 km 2 / K E) 2.5 x 10 6 km 2 / K 3. Suppose we know that the climate sensitivity " = 0.8 K / (W m -2 ). Further suppose that as the climate warms all the Arctic ice melts, preventing the ice-albedo feedback (f A = 0.30 W m -2 K -1 ) from operating. At the same time, a new methane-permafrost feedback starts (f M = 0.2 W m -2 K -1 ). What is the resulting climate sensitivity? A) 0.2 K / (W m -2 ) B) 0.34 K / (W m -2 ) C) 0.74 K / (W m -2 ) D) 1.25 K / (W m -2 ) E) 3.2 K / (W m -2 ) Use ࠵? = − 6 7 = −1 / ∑ ࠵? ; = in two steps, first find the total current feedback f c = -1/0.8 = -1.25 then subtract the old ice feedback, add the new methane feedback and turn it back into λ: ࠵? = -1/(-1.25 -0.3 + 0.2) = 0.74

EOSC 340 Quiz 2 practice problems 2022t1 2 /12 4. A black cloud layer at a height of 10 km is radiating 160 Wm -2 . Given an atmospheric lapse rate of -7 K/km, what is the best estimate for the temperature below the cloud at the Earth's surface? A) 290.5 K B) 295.5 K C) 300.5 K D) 305.5 K E) 310.5 K Two steps: 1. use equation to get the cloud temperature: (160./sigma)**0.25 = 230.48052309462736 2. Following the temperature profile down 10 km to the surface adds 70 degrees to give 300.5 K 5. According to the IPCC, the net radiative forcing by humans is currently between 1 and 2 Wm -2 above the pre-industrial value. If the climate sensitivity is 2.5 K/CO 2 doubling, what is the range of the equilibrium temperature increase between preindustrial times and today? A) between 0.42 K and 1.3 K B) between 0.44 K and 2.5 K C) between 0.65 K and 1.3 K D) between 1.36 K and 5 K E) between 2.45 K and 5 K 1 doubling is 3.8 Wm -2 = sensitivity = 2.5/3.8 = 0.66 K/(Wm -2 ) which is also the temperature change for 1 Wm -2 forcing. For 2 Wm -2 : 0.66*2 = 1.32 6. According to the IPCC, the magnitude of which of these climate feedback processes is the most uncertain? A) Planck feedback B) ice-albedo feedback C) lapse rate feedback D) cloud radiative feedback E) water vapour feedback 7. Let # s say we # re aiming to stabilize global temperature at 4 K above pre-industrial values. How many TOTAL Gtonnes of C can we emit if we are to achieve this goal, starting the count at the beginning of the industrial revolution? Assumptions to use: (1) assume climate sensitivity of 1 K (W m -2 ) -1 , (2) assume pre-industrial atmospheric CO 2 is 270 ppm, (3) assume 50% of what # s emitted stays in the atmosphere on the time scale of interest. A) about 300 Gtonnes carbon B) about 600 Gtonnes carbon C) about 1200 Gtonnes carbon D) about 2400 Gtonnes carbon E) about 4800 Gtonnes carbon Step 1, use Error! Reference source not found. to get the new CO 2 mixing ratio: np.exp(4/3.8*np.log(2))*270 = 560.06 Step 2: convert that to Gtonnes carbon assuming that half stays in the atmosphere: I = εσ T 4 Δ F = 3.8W m -2 ( ) ln newpCO2 / origpCO2 ( ) ln 2 ( )

EOSC 340 Quiz 2 practice problems 2022t1 3 /12 560 $ 270 = 290.0 2.1 * (560 - 270.)/0.5 = 1218.0 8. Which of the following greenhouse gases is a fast feedback, not a forcing? A) CO 2 B) H 2 O C) N 2 O D) CH 4 E) O 3 9. To obtain climate information about the past, we use oxygen isotopes measured in ice and also measured in the shells of marine organisms. Put the following in order from LIGHTEST ࠵? 18O to HEAVIEST ࠵? 18O value. i. ࠵? 18O of shells of deep sea benthic organisms living in the year 1750 ii. ࠵? 18O of shells of deep sea benthic organisms living at the last glacial maximum iii. ࠵? 18O of water in ice in Antarctica A. ii, i, iii B. iii, i, ii C. ii, iii, i D. iii, ii, i E. i, ii, iii Deep sea benthic shells will be heavier during LGM than today, mostly because of the ocean water will be enriched in (have more) 18O due to ice sheet growth. Ice in antarctica will consistently have very negative d 18 O due to strong fractionation during evap/transport/precip. 10. There is a strong relationship between local temperature and ࠵? 18O of precipitation today, as seen in this graph: What is the main/best explanation for this strong relationship? A. At colder temperatures, local evaporation takes up less O18, and so the water vapour in this region has less O18 – this is why precipitation in this region has lower ࠵? 18O, it is lighter.

EOSC 340 Quiz 2 practice problems 2022t1 4 /12 Note: it is true that local evaporation will have less O18 at colder temperatures, but this is not the MAIN cause of the relationship shown in the graph – assignment 4 takes you through why B is the correct answer B. Most water vapour is evaporated in the tropics. As the water vapour moves towards the poles, temperatures decrease, more and more water vapour condenses and falls as precipitation – because 18O is preferentially condensed, this causes the remaining water vapour to get lighter as the air travels further into colder regions, and thus precipitation also gets lighter. C. Most water vapour is evaporated in the tropics. As the water vapour moves towards the poles, temperatures decrease, more and more water vapour condenses and falls as precipitation – because 18O is preferentially condensed, this causes the remaining water vapour to get lighter as the air travels further into colder regions, and thus precipitation also gets lighter. D. Most water vapour is evaporated in the tropics. O16 is preferentially condensed and precipitated out, leaving more O18 by the time the water vapour reaches the poles. E. All water vapour has the same ࠵? 18O, but at colder temperatures more of the O16 condenses and is precipitated out, leading to more negative ࠵? 18O in the precipitation. 11. You measure a value of ࠵? 18O in an ice core of -20 % . Use the graph below to determine the residual vapour fraction of the air this precipitation came from: A. 0.01 B. 0.19 C. 0.4 - Remember that the ice core is recording the precipitation (rain/snow), not the water vapour itself. D. 0.8 12. Which of the following statements are accurate about paleoclimate data? i) Values of ࠵? 18O from ice cores give us very accurate estimates of average global temperatures by themselves - Not accurate – remember that the ࠵? 18O signal in ice cores can be complicated by changes in height of the ice sheet, changes in the seasonality of when the precipitation was mostly falling. ii) We use ice sheet borehole temperatures in combination with ࠵? 18O values of ice to get accurate temperatures