document 3

course phy 202

Reply from 02/05/07

you need to include a copy of that document, and mark additional work in some recognizable way (e.g,. use &&&&).

First I was trying to find P1. So I had (P1/P0)=(T1/T0). Then I had P1/1atm=300/270(270P1=300atm)/270 P1=1.11atm

pmax=patm[(tmax/to)-1]

10^5 Kpascals in 1 atm.

(1.11atm*10^5 - 10^5)=111000

(1000kg/m^3)(9.8m/s^2)('y)=11100

111000/9800='y

1.1326meters

(1.11atm*10^5 - 10^5)=11100 Pa

(1000kg/m^3)(9.8m/s^2)('y)=11100 Pa

111000 Pa / (9800 N / m^3) =`dy

`dy = 1.1326meters

(1000kg/m^3)(9.8m/s^2)(.2m)=1960Kpascals

At what temperature will that system give me that pressure?

101969/1atm=t2/270=275.92 would equal t2

v2/v1=t2/t1

v2/.5m^3=300C/275.292C

Everything looks good to this point.

(275.292v2=150)/275.292=v2=.54487M^3

'pg=mgh

.54487-.5=.04487m^3 volume pushed out

1000kg/m^3=m(kg)/v(m^3)

mass = 44.87 kg

44.87kg*9.8m/s^2*.2m=87.94 Joules

kg * m/s^2 * m gives you Joules. However you didn't get 44.89 kg into that bottle, much less raise that much water. That would be over 100 lbs of water.

"

**I'm not sure where you got your .5 m^3, which was apparently what you used as a basis for calculating volume change. My guess is that you meant .5 liters, which would be only .0005 m^3.**