Posted by amela on .
Carnot engine operates between 170°C and 23°C. How much ice can the engine melt from its exhaust after it has done 4.5 104 J of work?

PHYSICS PLEASE HELP DUE SOON 
PhysicsPro,
Post what you have so far, and I can guide you.

PHYSICS PLEASE HELP DUE SOON 
amela,
T1=273+170=443 K
T2=273+23=296 K
efficiency =(Q1Q2)/Q1 =W/Q1
efficiency =(T1T2)/T1
W/Q1 = (T1T2)/T1
Q1=W•T1/(T1T2)
Q1= r•m
r= 333•10⁵ J/kg
W•T1/(T1T2) =r•m
m = W•T1/(T1T2)• r =
(4.5x10^4)•443/(443296)•333•10⁵=4.5•10^12 
PHYSICS PLEASE HELP DUE SOON 
PhysicsPro,
A neat trick:
vbr{lr<WT1x^(1/9)><f(Q1Q2)/.bbn>
vbr<lr><5.12>(296^3)
.bbn=%eff^5
%ff=F(443)
.jrbn=%44x10^2
.qen=.133%tri
.yyn=.tri[kg]
t.r(333x10^5)i=vBr2^(1/3)
eff1=31.4%eff2
t4=59.3J/kg
t2=28.6J/kg
Tell me what you get: 
PHYSICS PLEASE HELP DUE SOON 
amela,
I am really not sure how to do this

PHYSICS PLEASE HELP DUE SOON 
amela,
I don't understand the abbreviations and symbols { <>, .yyn .jrbn etc

PHYSICS PLEASE HELP DUE SOON 
PhysicsPro,
What I did:
1)vbr functioneasiest to use
2).bbn is to solve .bbn is easy to the %eff^5
3)Set %ff to F(T1)
4)Make function of .jrbn
.jrbn=%continual change or %eff^5 which came out to 44x10^2
5)Function of .qen= 133% of tri or bbn final.
6)Function of .yyn=tri what you just solved for.
7)Rearrange tri which is equal to 1.33
1.3(333x10^5)i = vBr2^(1/3)
Solved for % of eff3
8)eff1=31.4%eff2self explanatory
9)then you get t4 and t2
10)Solve from that 
PHYSICS PLEASE HELP DUE SOON 
PhysicsPro,
If I didn't elaborate enough on step 7 you solve for vBr2^(1/3) for eff3

PHYSICS PLEASE HELP DUE SOON 
amela,
I get 2193.17

PHYSICS PLEASE HELP DUE SOON 
amela,
I give up