Question

Suppose you have an electric hot water heater for your house which is an aluminum cylinder which has a 0.5 m radius and is 1.4 m high. The walls are 1.0 cm thick. The thermal conductivity of Aluminum is 217 W/(m K). Assume that the temperature of the hot water inside the hot water heater is kept at a constant 90 C, and the external temperature is 27 C.

(See below for questions)

Answer 1

a) What is the surface area of the cylinder?
Express your answer using three significant figures.

>> Found answer: 2pi^2 2pi*r*h = 5.97 m^2

b) How much energy is lost through the walls of the hot water heater in one week? (Assume thinner surface of the heater is 90 C and the outer surface is 27 C.)

>>> Trying to use equation Q/delta T = kA(delta T/L) but getting a large, wrong number. Units are in joules.

c) Assume you pay $0.10 per kW-hour for electricity. How much would it cost just to keep the hot water inside the heater for one week? Unit is in $.

d) Suppose that you wrap the hot water heater on all sides with a 10 cm thick blanket of fiberglass insulation which has a thermal conductivity of 0.04 W/(m K). Assume the inner surface of the fiberglass insulation is at 90 C and the outer surface is at 27 C, and the total surface area is still what you calculated in part A. Units is in J.

e) Assume you pay $0.10 per kW-hour for electricity. How much would it cost just to keep the hot water inside the fiberglass-wrapped heater for one week? Units is in $.

Answer 2

Not sure, have you tried looking at your book for some examples?

Answer 3

b) Q/delta T = kA(delta T/L)
(217)(5.97)(63/0.01) = 8161587 Joules
joules = watts * second = 8161587 * 604800 = 4.9*10^12

Answer 4

I'm stuck on c)
I tried....

10 cents per kW 8161587 watts per week >>> 8161.587 kw x 0.1 = $816.16 Which was wrong.

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4.9 * 10^12 * 7 days * 24 hours * 60 minutes * 60 seconds = 2.96 * 10^18
2.96 * 10^18 / 3600000 kW per hour = 8.23 * 10 ^ 11* 0.1 m = 8.2 * 10^10 dollars?

Obviously that's way too big...

Answer 5

d) is the same as b) and I got 9.10×107 J, which was right.

So I'm stuck on the process of c) and e)

Answer 6

I tried...
kW = J/1000*seconds per week
=4.9*10^12/1000*604800
=8101.85 * 0.1 = $810.19

Wrong.
I'm missing something here...

Answer 7

I think the issue is that you are doing calculations based on 10 cents per kW and the question reads 10 cents per kW-hr. Check out the conversion here.
http://en.wikipedia.org/wiki/Kilowatt_hour

Answer 8

Wow. That made it 10 times easier. I didn't realize it was a unit on it's own.
I really wish my teacher or the book explained that unit, because that's misleading x.x

Answer 9

So 1.37*10^6 kW-hr is 3.7*10^5... okay let me see if I got e) too

Answer 10

Glad to help out. :) Yeah, it's nice to know when whacky units exist...

Answer 11

Yeah , $2.53 for the fiberglass one. I got it right.

Gosh, I was working WAY to hard for this haha. But now I know.

Answer 12

Yup, that's what I get. Great job!