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?
Post what you have so far, and I can guide you.
T1=273+170=443 K
T2=273+23=296 K
efficiency =(Q1-Q2)/Q1 =W/Q1
efficiency =(T1-T2)/T1
W/Q1 = (T1-T2)/T1
Q1=W•T1/(T1-T2)
Q1= r•m
r= 333•10⁵ J/kg
W•T1/(T1-T2) =r•m
m = W•T1/(T1-T2)• r =
(4.5x10^4)•443/(443-296)•333•10⁵=4.5•10^12
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:
I am really not sure how to do this
I don't understand the abbreviations and symbols { <>, .yyn .jrbn etc
What I did:
1)vbr function--easiest 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%eff2--self explanatory
9)then you get t4 and t2
10)Solve from that
If I didn't elaborate enough on step 7 you solve for vBr2^(1/3) for eff3
I get 2193.17
I give up
To answer this question, we need to utilize the principles of the Carnot cycle and energy conservation.
The Carnot engine is a theoretical engine that operates between two heat reservoirs at different temperatures. It is known for its maximum efficiency, which is given by the equation:
Efficiency = (T1 - T2) / T1
Where T1 is the temperature of the hot reservoir and T2 is the temperature of the cold reservoir. In this case, T1 is 170°C and T2 is 23°C. We can convert both temperatures to Kelvin by adding 273:
T1 = 170 + 273 = 443 K
T2 = 23 + 273 = 296 K
Now we can calculate the efficiency:
Efficiency = (443 - 296) / 443 = 0.335
The efficiency of the Carnot engine tells us how much of the input energy is converted into useful work. In this case, the engine has done 4.5 * 10^4 J of work. To find out how much energy was supplied to the engine, we can use the equation:
Work = Efficiency * Input Energy
Input Energy = Work / Efficiency
= (4.5 * 10^4 J) / 0.335
≈ 1.343 * 10^5 J
Now, let's move on to the ice melting. For this, we need to know the latent heat of fusion of water, which is the amount of energy required to melt one gram of ice. The latent heat of fusion of water is approximately 334 J/g.
Using this value, we can calculate the amount of ice melted by dividing the input energy by the latent heat of fusion:
Amount of ice melted = Input Energy / Latent Heat of Fusion
≈ (1.343 * 10^5 J) / (334 J/g)
To convert the mass of ice melted into grams, we need to know the density of ice. The density of ice is approximately 0.92 g/cm^3.
Let's assume the ice is in the form of a cube with length L. The volume of the cube is L^3, and since the density is 0.92 g/cm^3, the mass of the cube is 0.92 * L^3 g.
So, the amount of ice melted is approximately:
Amount of ice melted ≈ (1.343 * 10^5 J) / (334 J/g) * (1 / (0.92 * L^3))
To find the value of L, we would need additional information about the engine or its exhaust.