please answer number 11 Outlook Web × i P142 HW1 x M what mass of x \\ G How muc
ID: 1661509 • Letter: P
Question
please answer number 11
Outlook Web × i P142 HW1 x M what mass of x G How much he × Sgn into XFir × D self-service- x-p PaperCut MF x G specific heat) d www.webassign.net/web/Student/Assignment-Responses/last?dep: 18090635 - × According to our records you have not yet redeemed an access code for this class or purchased access online. The grace period will end Tuesday, January 30 2018 at 12:00 AM PST. After that date you will no longer be able to see your WebAssign assignments or grades Get access now, 10. 0/1 pointsI Previous Answers Giancoli? 14.P.025 My Notes Ask Your Teach How much heat is needed to melt 17.90 kg of silver that is initially at 17°C? 7.3e+06 0/1 points | Previous Answers Giancoli7 14.P.027 My Notes Ask Your Teach What mass of steam at 100°C must be added to 1.02 kg of ice at 0°C to yield liquid water at 21°C? 2098 % kg Submit AnswerSave Practice Another Version 12. -11 points Giancoli7 14.P.053. My Notes Ask Your Teach On a hot day's race, a bicyclist consumes 7.6 L of water over the span of four hours. Making the approximation that all of the cyclist's energy goes into evaporating this water as sweat, how much energy in kcal did the rider use during the ride? (Since the efficiency of the rider is only about 15 percent, most of the energy consumed does go to heat, so our approximation is not far off.) Use 585 kcal/kg for the latent heat of vaporization of water at room temperature kcal add-drop-revised.-.. af Show all | × 9:02 AM ^qx 1/21/2018 |Explanation / Answer
11)
Heat out of the steam equals heat into ice
Q out is in two steps
first step condensing the steam to water using Q = m*Lv = m*2.256x10^6J/kg
Step two lowering temp from 100 to 21 using Q = m*c*deltaT = m*4184J/kgC*79 = m*330536
Adding these we get 2.59x10^6m
Q into ice is in two steps also
Step 1 melt the ice Using Q = m*Lf = 1.02kg*3.34x10^5J/kg = 3.4x10^5J
Step 2 Increase T to 21 from 0 Q =m*c*deltaT = 1.02*4184*21 = 89621.28J
Total = 2.92376x10^6J
Equating these give 2.59x10^6m = 429621.28
Hence m steam = 429621.28/2.59x10^6 = 0.165kg
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.