kean (msk955) – Homework 10 – yao – (56725) 1This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsThe coldest temperature possible is −273 de-grees Celsius, which is called absolute zero.What is this temperature in degreesFahrenheit?1. −523.4◦F2. +491.4◦F3. −119.667◦F4. 32◦F5. +459.4◦F6. 0◦F7. +523.4◦F8. −491.4◦F9. −32◦F10. −459.4◦F correctExplanation:Let : TC= −273◦C .TF=95TC+ 32 =95(−273◦C) + 32=−459.4◦F .002 10.0 pointsWhat is the boiling point of ether on theKelvin scale? The boiling point of ether is35◦C.1. 35 K2. −238 K3. 208 K4. 308 K correctExplanation:Tk= 35◦C + 273 =308 K .003 10.0 pointsAt what temperature is the Celsius scale read-ing twice the Fahrenheit scal e reading?1. −20◦F2. −24.6◦F3. −12.3◦F correct4. −12.3◦C5. −6.1◦CExplanation:Let : X = TFandTC= 2 X .TC=59(TF− 32) , so2 X =59(X − 32)2 X =59X −1609139X =−1609X =−16013= −12.3077 ,so TF= −12.3077◦C .004 10.0 pointsYou need to raise the temperature of 57 g ofwa ter from 10.4◦C to 90◦C.How much heat is needed to accom-plish this? The specific heat of water is4180 J/kg ·◦C.Correct answer: 18965.5 J.Explanation:kean (msk955) – Homework 10 – yao – (56725) 2Let : m = 57 g = 0.057 kg ,c = 4180 J/kg ·◦C ,Tf= 90◦C , andTo= 10.4◦C .The temperature difference in the Cent igradescale is the same as the temperature di fferencein the Kelvin scale, so the heat absorb ed mustbeQ = m c ∆T = m c (Tf− To)= (0.057 kg) (4180 J/kg ·◦C)× (90◦C − 10.4◦C)=18965.5 J .005 10.0 pointsCup A contains 100 grams of water at 0◦Cand cup B contains 200 gr a m s of water at 50◦C. The contents of the two cups are mixedtogether in an insulated container (no heatcan leak in or out).What is the final temperature of the waterin the container?1. Between 25◦C and 50◦C correct2. Higher than 50◦C3. 50◦C4. 25◦C5. Lower than 0◦C6. Between 0◦C and 25◦CExplanation:Let : mA= 100 g ,TA= 0◦C ,mB= 200 g , a ndTA= 50◦C .Q = m c ∆T , so ∆T ∝1m.∆TA∆TB= −mBmA= −2T − TA= −2 (T − TB)3 T = TA+ 2 TBT =TA+ 2 TB3=0◦C + 2 (50◦C)3= 33.3333◦C .006 10.0 pointsThe air temperature above the coastal areasis profoundly influenced by the large sp ecificheat of water. One reason is that the heatreleased when 3 m3of water cools by 3◦C willraise the temperature of an enormously largervolume of air by 1◦C.Calculate this volume of air. The specificheat of air is approximately 1 kJ/kg ·◦C andits density is 1.25 kg/m3. The specific heatof water is 4.19 kJ/kg ·◦C and its density is1000 kg/m3.Correct answer: 30168 m3.Explanation:Let : Vw= 3 m3,cw= 4.19 kJ/kg ·◦C ,ρw= 1000 kg/m3,cair= 1 kJ/kg ·◦C ,ρair= 1.25 kg/ m3,∆Tw= 3◦C , and∆Tair= 1◦C .ρwVwcw∆Tw= ρairVaircair∆TairVair=ρwVwcw∆Twρaircair∆Tair=(1000 kg /m3)(3 m3)1.25 k g/m3×(4.19 kJ/kg ·◦C) (3◦C)(1 kJ/kg ·◦C) (1◦C)= 30168 m3.kean (msk955) – Homework 10 – yao – (56725) 3007 10.0 points400 g of wa ter at 99◦C is po ured i nto a 18 galuminum cup containing 70 g of water at11◦C.What is the equilibrium temp erature ofthe system? The specific heat of wa-ter is 1 cal/g ·◦C and of aluminum is0.215 cal /g ·◦C.Correct answer: 85.282◦C.Explanation:Let : mh= 400 g ,mc= 70 g ,mAl= 18 g ,cw= 1 cal/g ·◦C ,cAl= 0.215 cal/g ·◦C ,Tc= 11◦C , andTh= 99◦C .The heat lo st by hot water equals the heatgained by cold meta l plus the water, somhcw(Th− Tf)= mAlcAl(Tf− Tc) + mccw(Tf− Tc)(mhcw+ mAlcAl+ mccw) Tf= mhcwTf+ mAlcAlTc+ mccwTcTf=mhcwTh+ mAlcAlTc+ mccwTcmhcw+ mAlcAl+ mccwSincemhcwTh+ mAlcAlTc+ mccwTc= (400 g) (1 cal/g ·◦C) (99◦C)+ (18 g) (0.215 cal/ g ·◦C) (11◦C)+ (70 g) (1 cal/g ·◦C) (11◦C)= 40412.6 cal andmhcw+ mAlcAl+ mccw= (400 g) (1 cal /g ·◦C)+ (18 g) ( 0.215 cal/g ·◦C)+ (70 g) ( 1 cal/ g ·◦C)= 473.87 cal/◦C ,then the equilibrium temperature will beTf=40412.6 cal473.87 cal/◦C=85.282◦C .008 10.0 pointsA cowboy fires a silver bullet of mass 3 g witha muzzle speed of 188 m/s into the pine wallof a saloon.What is the t emperature change of the bul-let? Assume t hat all the internal energy gen-erated by the impact remains with the bullet.The specific heat of silver is 234 J/kg ·◦C.Correct answer: 75.5214◦C.Explanation:Let : m = 3 g ,v = 188 m/s , andc = 234 J/kg ·◦C .The kinetic energy of the bullet isEk=12m v2=12(0.003 kg) ( 188 m/s)2= 53.016 J .Nothing in the environment is hotter than thebullet, so the bullet gains no thermal energy.Its temperature i ncreases because the 53.016 Jof kinetic energy became 53.016 J of extrainternal energy. ThusEk= Q = m c ∆T∆T =Qm c=53.016 J(0.003 kg) (234 J/kg ·◦C)=75.5214◦C .009 10.0 pointsWhy are feather beds warm and why is goosedown considered the best filling for a parka?1. Goose down and feathers have high spe-cific heat.2. Goose down and feathers have a highernatural temperature than other materi als.3. Goose down and feathers trap a l ot of air,which is a good insulator. correctExplanation:kean (msk955) – Homework 10 – yao – (56725) 4Both feathers and goose down trap air,which is a good insulator.010 10.0 pointsThe radiation per unit area from the Sunreaching the earth is 1400 W/m2, approx-imately the amount of radiative power perunit area reaching a sun bather o n the banksof Barton Springs at noon on a clear day inJune. The temperature of the Sun is 580 0 K.Now suppose instead of the present Sun wereceived radiati on from sun X at temperature2863 K, l ocated at the same position as ourSun.How much radiative power per unit areawo ul d reach the sun bather from the new sunX?Correct answer: 83.1192 W/m2.Explanation:Let : P1= 1400 W/m2,T1= 5800 K , andT2= 2863 K .P ∝ T4for a given area, soP2P1=T2T14P2=T2T14P1=2863 K5800 K4(1400 W/m2)=83.1192 W/m2.011 10.0 pointsAn unclothed student is in 20◦C room.If the skin temperature o f the student is37◦C, how much heat is lost from his b odyin 17 min, assuming that the emissivity of theskin is 0.82 and the surface area of the studentis 1.3 m2. The St ephan-Boltzmann constantis 5.6696 × 10−8W/m2· K4.Correct answer: 1.14981 × 105J.Explanation:Let : A = 1.3 m2,σ = 5.6696 × 10−8W/m2· K4,e = 0.82 ,Tstudent= 37◦C = 310 K ,Troom= 20◦C = 293 K , andt = 17 min .Using Stefan’s law P = σ A e T4, the ra teof thermal energy loss from the skin