# Students` name Topic

Final Exam Spring 2015

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Question 1

Soda:

m=0.34kg

c=4190

initialtemp= 61 degrees

finaltemp= 6 degrees

change=(61-6)= 55 degrees

Ice:

m=?

c=2000

initialtemp=-19 degrees

finaltemp= 6 degrees

change=(6-(-19))=25 degrees

working:

(m*c*changein temperature of soda)=(m*c*change in temperature of ice )

(0.34*4190*55)=(x*2000*25)

78353=50000x

x=78353/50000

x=1567g

coolingthe can can:

m=0.02kg

c=900

changein temperature= (61-6)= 55 degrees

working:

(m*c*changein temp of can)+(m*c*change in temp of soda)=(m*c*change in temp ofice)

(0.02*900*55)+(0.34*4190*55)=(x*200*25)

990+78353=50000x

x=79343/50000

x=1587g

massof ice needed = (1587-1567)g

=20grams

Question2:

N=15

vol=1litre

temp=293k

d=1cm

m=2g

Meanfree path of jelly beans

=1/√2*π*d2*N/v

=1/√2*π*0.012*15/0.001m3

=1/6.6643

=0.15m

RMSvelocity

=√(3RT/M)

r=8.3145

t=293k

m=0.002kg

=√(3*8.3145*293/0.002)

=1911.66m/s

Timebetween collisions:

t=d/v

d=0.15m

v=1911.66m/s

t=0.15/1911.66

t=7.847*10-5 seconds

Pressureexerted by a jelly bean on the container floor

P=F/A

F=0.002*9.807

A=π*0.0052

F=0.002*9.807/π*0.0052

F=249.73pascals, a jelly bean does not exert pressure on the walls of the containerbecause due to its solid state and its weight the pressure it exertsis downwards

MeanFree path of Nitrogen

=1/√2*π*d2*N/v

=1/√2*π*(2*10-10m)2*15/0.001m3

=1/4.443*6*10-16m

=3.75*1014m

RMSvelocity of Nitrogen

=√(3RT/M)

r=8.3145

t=293k

m=4.68×10 -26 kg

=(√3*8.315*293/4.68×10-26)

=3.952*10-12m/s

pressureexerted by the gas

PV=nRT

pressure=nRT/v

p=pressure

v=0.001m3

R=8.315

T=293k

n=15mols

P=15*8.315*293/0.001

P=36544.425/0.001

P=36543236.153 Pascals

Timebetween collisions:

t=d/v

d=3.75*1014m

v=3.952*10-12m/s

t=3.75*1014m/3.952*10-12m/s

t=95sec

Question3:

n= (TH-TC)/TH

TH=373.15k

TC=273.15k

n=(373.15-273.15)/373.15

n= 26.8%

Therefore, not possible for the efficiency to be 75%

n= (TH-TC)/TH

n=(373.15-273.15)/373.15

n= 26.8%

QH

e=(QH-QC)/QH

w= QH-QC

0.268=1200/ QH

QH=1200/0.268

QH=4477.6J

QC=QH -w

QC=4477.6-1200

QC=3277.6J

W=QH-QC

= 4477.6-3277.6

=1200J

Question 4:

At point 1

Pressure is 1atm and volume is 0.001m3

point 1 to 2 =Isochronic

At point 2

Pressure is at 5atm and volume is 0.001m3

point 1 to 3= Isobaric

At point 3

Pressure is at 1atm and volume is Vmax

Temperature at 2 and 3 is thesame(isothermal)

n= (TH-TC)/TH

PV=nRT

pressure=nRT/v

T1=1.015*105*1*10-3/0.03*8.3

=407.63k

T2=5* 1.015*103*407.63/1.015*103

=2038.15k

efficiency=W/Q

w=nRTIn(Vf/Vi)

efficiency=0.03*8.325*2038.15In(5/1)

816.78J

Q1= nCv*change in temperature

=0.03*12.5*(2030-407.6)

=611.025J

Q2=0.03*20.8*1629.4

=1016.745

n=816.78 J/1016.745

=50.17

nca=1-tm/tf

=1-407.63/2038.15

=1-0.2

=0.8

Theengine is real

Question5:

Q1T1=Q2T2

Q1=(Q2T2) /T1

Q3T3=Q4T4

Q3=(Q4T4) /T3

Q3/Q1=(Q4T4/T3) /(Q2T2/T1)

WhereQ4=Q2 therefore:

Q3/Q1=(T4T1 )/(T3T2)

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