Heat problem (cooler at a beach)

  • Thread starter goonking
  • Start date
  • Tags
    Heat
In summary: The answer was 1.57 kgOk. 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 average weight of an ice cube is about 0.9 grams. So Q would be about 2.1 kilograms.The answer was 1.57 kgOk. 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 average weight of an ice cube is about 0.9 grams
  • #1
goonking
434
3

Homework Statement


upload_2015-5-15_1-40-23.png


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?
 

Attachments

  • upload_2015-5-15_1-39-30.png
    upload_2015-5-15_1-39-30.png
    8.7 KB · Views: 419
Last edited:
Physics news on Phys.org
  • #2
goonking said:

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.
 
  • #3
Raghav Gupta said:
Hey, that L is not latent heat in equation but length.
oh, sorry! good catch
 
  • #4
goonking said:
oh, sorry! good catch
There is another equation for heat transfer what is that?
 
  • #5
Raghav Gupta said:
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?
 
  • #6
goonking said:
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?
 
  • #7
Raghav Gupta said:
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?
 
  • #8
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.
 
  • Like
Likes goonking
  • #9
In case it's not obvious... In this case L is the thickness of the insulation (eg the length of the heat path)
 
  • Like
Likes goonking
  • #10
Raghav Gupta said:
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?
 
  • #11
goonking said:
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.
 
  • #12
goonking said:
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).
 
  • #13
Yeah, sorry.
The length to be taken is 2cm.
 
  • #14
CWatters said:
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
 
  • #15
goonking said:
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?
 
  • Like
Likes goonking
  • #16
CWatters said:
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
 
  • #17
Yes. I make the equation..

Q(Joules) = (0.035) (6 * 0.4m x 0.4m) (10800 seconds) (29 C) / (0.2m)
 
  • Like
Likes goonking
  • #18
CWatters said:
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
 
  • #19
Yes you are right.
 
  • #20
Yes sorry /0.02.
 
  • #21
Raghav Gupta said:
It is 0.157 kg.
make sure you put in 0.02 m, not 0.2m
 
  • #22
Yes, sorry,
Edited.
 
  • #23
Just a thought but..

The question isn't clear if 0.4m is the inside or outside dimension of the box.
 
  • #24
CWatters said:
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
 
  • #25
goonking said:
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?
 
  • #26
Raghav Gupta said:
Isn't the answer given in your problem so you could verify?
yes, the answer was 1.57 kg
 
  • #27
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.
 
  • Like
Likes goonking

FAQ: Heat problem (cooler at a beach)

How does a cooler keep drinks cold at the beach?

A cooler works by insulation and cooling mechanisms. The insulation keeps the cool air inside the cooler from escaping, while the cooling mechanisms, such as ice or gel packs, lower the temperature inside the cooler.

Why does a cooler work better in the shade at the beach?

A cooler works better in the shade because it is not exposed to direct sunlight, which can increase the temperature inside the cooler. Being in the shade also helps to keep the cooler's insulation and cooling mechanisms from melting or warming up too quickly.

How long will a cooler keep drinks cold at the beach?

The length of time a cooler can keep drinks cold at the beach depends on various factors, such as the outside temperature, the amount and type of insulation, and the type and amount of cooling mechanisms. In general, a well-insulated cooler with enough ice or gel packs can keep drinks cold for several hours at the beach.

Can I use a regular cooler at the beach or do I need a specialized beach cooler?

You can use a regular cooler at the beach as long as it has enough insulation and cooling mechanisms to keep your drinks cold. However, specialized beach coolers often have additional features, such as sand-resistant exteriors and built-in cup holders, that can make them more convenient for beach use.

How can I make my cooler more effective at keeping drinks cold at the beach?

To make your cooler more effective at keeping drinks cold at the beach, you can use a combination of techniques. These include pre-chilling the cooler and its contents, using more insulation, such as towels or blankets, and using larger and more efficient cooling mechanisms, such as dry ice or frozen water bottles.

Similar threads

Replies
4
Views
3K
Replies
1
Views
928
Replies
5
Views
1K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
7
Views
3K
Back
Top