Heat problem (cooler at a beach)

[goonking]

Homework Statement

Homework Equations

Q = kAΔT/L

The Attempt at a Solution

[/B]
so I need to find the volume of the cooler, and then the volume is the volume is ice in it?

then I need to find the surface area of 1 side of the cooler, than multiply it by 6. then plugging in the total area in Q = kAΔT/L

is this a good approach to the problem?

 

[Raghav Gupta]

Homework Statement

View attachment 83505

Homework Equations

Q = kAΔT/L (L = latent heat)

The Attempt at a Solution

[/B]
so I need to find the volume of the cooler, and then the volume is the volume is ice in it?

then I need to find the surface area of 1 side of the cooler, than multiply it by 6. then plugging in the total area in Q = kAΔT/L

is this a good approach to the problem?

Hey, that L is not latent heat in equation but length.

 

[goonking]

Hey, that L is not latent heat in equation but length.

oh, sorry! good catch

 

[Raghav Gupta]

oh, sorry! good catch

There is another equation for heat transfer what is that?

 

[goonking]

There is another equation for heat transfer what is that?

Q = L m

we equate them so we get

L m = kAΔT/Length

we have L

we have k

we have A and L

I'm assuming ΔT is 29 C?

 

[Raghav Gupta]

Q = L m

we equate them so we get

L m = kAΔT/Length

we have L

we have k

we have A and L

I'm assuming ΔT is 29 C?

How to bring time term here which is 3 hrs?

 

[goonking]

How to bring time term here which is 3 hrs?

i think the answer we get, if we did all the algebra, is the amount of heat per seconds.

so we just convert 3 hours to seconds. then multiply by the heat per seconds

correct?

 

[Raghav Gupta]

First of all the formula for heat conduction is
Q = kAtΔT/L where k is constant , A is area, t is time, ΔT is temperature difference and L is length.

 

[CWatters]

In case it's not obvious... In this case L is the thickness of the insulation (eg the length of the heat path)

 

[goonking]

First of all the formula for heat conduction is
Q = kAtΔT/L where k is constant , A is area, t is time, ΔT is temperature difference and L is length.

Q = (0.035) (0.02m x 0.4m) (10800 seconds) (29 C) / (0.4m) = 219.24 joules

i believe we set Q = L m = 219.24 joules

m should be 6.56x10^-4kg

correct?

 

[Raghav Gupta]

Q = (0.035) (0.02m x 0.4m) (10800 seconds) (29 C) / (0.4m) = 219.24 joules

i believe we set Q = L m = 219.24 joules

m should be 6.56x10^-4kg

correct?

Yes, it is correct.

 

[CWatters]

Q = (0.035) (0.02m x 0.4m) (10800 seconds) (29 C) / (0.4m) = 219.24 joules

No "A" is the surface area. You were right when you wrote..

I need to find the surface area of 1 side of the cooler, than multiply it by 6. then plugging in the total area in Q = kAΔT/L

L is the length of the heat path which is through the walls (eg 2cm).

 

[Raghav Gupta]

Yeah, sorry.
The length to be taken is 2cm.

 

[goonking]

No "A" is the surface area. You were right when you wrote..
L is the length of the heat path which is through the walls (eg 2cm).

oh, i forgot to multiply the area by 6, so it should be 6( 0.02m x 0.4m) = A

so new answer should be 26300 joules

i would have never thought the length was 2cm

 

[CWatters]

i would have never thought the length was 2cm

kAΔT/L gives you the power (energy per second) flowing through the walls. If the walls are thicker (better insulated) then L is longer and less power flows. Make sense?

 

[goonking]

kAΔT/L gives you the power (energy per second) flowing through the walls. If the walls are thicker (better insulated) then L is longer and less power flows. Make sense?

ok, got the answer to be 1.57 kg, sounds much more reasonable now

 

[CWatters]

Yes. I make the equation..

Q(Joules) = (0.035) (6 * 0.4m x 0.4m) (10800 seconds) (29 C) / (0.2m)

 

[goonking]

Yes. I make the equation..

Q(Joules) = (0.035) (6 * 0.4m x 0.4m) (10800 seconds) (29 C) / (0.2m)

should be 0.02 m

 

[Raghav Gupta]

Yes you are right.

 

[CWatters]

Yes sorry /0.02.

 

[goonking]

It is 0.157 kg.

make sure you put in 0.02 m, not 0.2m

 

[Raghav Gupta]

Yes, sorry,
Edited.

 

[CWatters]

Just a thought but..

The question isn't clear if 0.4m is the inside or outside dimension of the box.

 

[goonking]

Just a thought but..

The question isn't clear if 0.4m is the inside or outside dimension of the box.

true, i was thinking that too when I tried to draw out the cooler

 

[Raghav Gupta]

true, i was thinking that too when I tried to draw out the cooler

Isn't the answer given in your problem so you could verify?

 

[goonking]

Isn't the answer given in your problem so you could verify?

yes, the answer was 1.57 kg

 

[CWatters]

Ok. Just for info.. If you assume they meant the outside dimension then the insides are about 0.36 * 0.36. If you use that for the area the result is about 1.28kg. Real world the answer will be somewhere in between the two.