Question

The pot is in the shape of a round cylinder made of aluminum which is 10 cm in radius. The aluminum is 3 mm thick. You notice that after 11 minutes, 0.13 kg of water has boiled away.

I already did a). I found the inner surface of bottom of the pot to be 100 C.

b) How much energy is required to boil away M kg of water?
Answer is in units of J.

c) What is the difference in temperature between the inner and outer sufaces of the aluminum bottom of the pot? Answer is in units of C.

d) What is the change in entropy of the boiling water after boiling away 0.13 kg of water?
Answer is in units of J/K.

Answer 1

is the 10 cm the inner or outer radius?

Answer 2

Energy needed can be found from latent heat of vaporization, 539 cal/gm,
convert to kg and J.

Heat transfer rate is is dQ/dt = k A (T1-T2)/L where k is conductivity of aluminum. You know dQ/dt from rate at which you boiled off the water.
L is thickness of pan, A the area. T2=100 oC. Solve for T1

Entropy change is related to Q/T, but may be Q/(1/T2-1/T1) and should be >0.

Answer 3

@roadjester ; I think it just means if you are standing above it, the pot is 10cm in radius as a whole.

@douglaswinslowcooper ; Lemme work it out with this information and see if I can find the right answers. thanks for the info.

Answer 4

b) 539 cal/g = 2256 kJ/kg = 2.26 × 10^6 J/kg or 2256000 ; It said this was wrong.
c) Didn't do this yet since I need b).
d) Didn't do this since I need c).

Answer 5

"b) How much energy is required to boil away M kg of water?"
=Latent Heat of M kg of water

"c) What is the difference in temperature between the inner and outer sufaces of the aluminum bottom of the pot? Answer is in units of C."
Assume steady state and find the temperature difference required for "...after 11 minutes, 0.13 kg of water has boiled away..."

PS: I am still trying to find a way to solve entropy. Q/(1/T2-1/T1) may cause trouble as T1=T2 (temperature during boiling remains constant)

Answer 6

b) Okay, so 22.6*10^5 J/kg is the latent heat of vaporization according to my book. No one really defined M, so I assumed it meant 1 mole of water, which meant 18g. I converted and used L = Q/m to find Q, and still did not get the right answer. The correct answer was 2.94×10^5 J.

Answer 7

c) dQ/dt = kA(T1-T2)/L
dQ/dt = 2.94*10^5 ; k=217; A= ?; T1 = ?; T2 = 100 C

So, how am I suppose to find A?
I know that A = (2)(pi)(r)^2 + (2)(pi)(r)(h) of a cylinder.
But the problem only gave me the radius and not the height?

Answer 8

d) Q = mL = (0.13) (22.6*10^5) = 293800 J
Change in Entropy = Q/T = 293800/Part C

^ I think that's how to do it, but I have no idea what Part C is yet.

Answer 9

Part (b)
Answer depends on the value of M (try 0.13 kg or complete volume of water).

Part (c)
You need to find delta T
A = surface area of pot

Part (d)
"...Change in Entropy = Q/T ..."
I don't really see how?

PS: I am not checking your calculations; only the logical consistency of your solution (numbers gimme a headache :P )

Answer 10

M is 0.13 kg of water.
The pot is so thin that the difference in radius for the bottom of 3mm is negligible. Use the round number. The heat is coming through that area, much less so from the side, so ignore side, and you don't know the temperature of the side for sure anyway. Heat transport issue was not finding A but finding the difference between the bottom temperature of the pot (T2) and the inner temperature (T1=100oC).

Answer 11

dQ/dt = kA(T1-T2)/L
dQ/dt = 2.94*10^5 ; k=217; A= ... ; T1 = ?; T2 = 100 C; L = 0.003m

Okay, so I tried using 0.1m as the area and then (2)(pi)(0.1)^2 = 0.0628 as the area.
I got the answers of 35.28 and 59.35 C respectively for T2 but neither were right.

Answer 12

Make sure you are in units that are consistent, SI: m, kg, s etc.
Pot bottom surface area should be pi (r)^2 where r = 0.1m, area of a disk.
Make sure your units are consistent.

Answer 13

(pi)(0.1 m)^2 = 0.03
2.94*10^5 J = (217 W/m*k)(0.03 m)(100 - T1) / 0.003 m
T1 = 35.5

SI units are consistent, and this is very close to my first try above, and it's still telling me I'm wrong.

Answer 14

area (pi) (0.1)^2 = 0.0314 m2

(b) heat needed = (0.13 kg)(2.26 x10^6 J/kg) = 2.94 x 10^5 J yes

(c) rate averaged over 11 minutes = (2.94 x 10^5 J)/(11)(60)=445 J/s or W

rate = k A dt/dx = (238 W/m oC) (0.0314 m2) (T-100 oC)/(0.003 m)

445 W = (238 W/m oC) (0.0314 m2) (T-100 oC)/(0.003 m) units check

(T - 100) = (445)(0.003)/(238)(0.0314) = 0.179 oC

T = 100.179 oC

Seems like an awfully small temperature difference, but it is a big thin pot.???

Answer 15

Your last attempt used the total heat rather than the heat transfer rate dQ/dt = Q/(11min)(60 s/min) = 445 J/s, and we had slightly different values for k for aluminum.

Answer 16

Yeah, exact number was 0.196

Let's see if I can figure out d) now...

Answer 17

It needs to be in J/K so the only think I could figure that made sense was (2.94*10^5 J)/(0.196 K) to get those units, but 15000000 J/K, which was obviously wrong.

Answer 18

Number needs to be small amount above 100 oC.

Answer 19

Eh... 150?
I only have one try left, so I don't want to be wrong and drop 25% on a question...

Answer 20

Or do you something small above 100, more like 100.179?