Temperature change with sliding a crate

In summary: I hope you remember this bit of calculator operations for the future.:biggrin: Glad to help. I hope you remember this bit of calculator operations for the future.In summary, a crate of fruit with a mass of 33.5kg and a specific heat capacity of 3650 J/kg*K slides 8.20m down a ramp inclined at an angle of 37.6 degrees below the horizontal. Using the equation Change in temperature = gLsin(theta)/c - v^2/2c, and given the speed of 2.4 m/s at the bottom of the incline, the temperature change is calculated to be approximately 0.01 degrees Celsius. Remember to use radians instead of degrees
  • #1
Bigworldjust
54
0

Homework Statement


A crate of fruit with a mass of 33.5kg and a specific heat capacity of 3650 J/kg*K slides 8.20m down a ramp inclined at an angle of 37.6 degrees below the horizontal.

If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change ?

Homework Equations



Change in temperature = gLsin(theta)/c - v^2/2c


The Attempt at a Solution



I am doing the following equation above and I am getting an answer of -3.00 * 10^-3 and it keeps saying it is wrong. I am not sure what I am doing wrong. I have it set up like this:

9.81*8.2*sin(37.6)/3650 - (2.4)^2/(2)(3650)

Any help would be appreciated, thanks!
 
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  • #2
Bigworldjust said:

Homework Statement


A crate of fruit with a mass of 33.5kg and a specific heat capacity of 3650 J/kg*K slides 8.20m down a ramp inclined at an angle of 37.6 degrees below the horizontal.

If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change ?

Homework Equations



Change in temperature = gLsin(theta)/c - v^2/2c


The Attempt at a Solution



I am doing the following equation above and I am getting an answer of -3.00 * 10^-3 and it keeps saying it is wrong. I am not sure what I am doing wrong. I have it set up like this:

9.81*8.2*sin(37.6)/3650 - (2.4)^2/(2)(3650)

Any help would be appreciated, thanks!

Where did the number 2.4 come from? Are you sure that your problem description is complete?
 
  • #3
gneill said:
Where did the number 2.4 come from? Are you sure that your problem description is complete?

Oh sorry that was given in the previous question as its speed:

If the crate was at rest at the top of the incline and has a speed of 2.40 m/s at the bottom
 
  • #4
Bigworldjust said:
Oh sorry that was given in the previous question as its speed:

If the crate was at rest at the top of the incline and has a speed of 2.40 m/s at the bottom

Okay. Well, your calculation method looks okay then. Check that your calculator is set for degrees and not radians for angle computations.
 
  • #5
gneill said:
Okay. Well, your calculation method looks okay then. Check that your calculator is set for degrees and not radians for angle computations.

Yeah, I did, and it still says its wrong. I got this as a suggestion on MasterPhysics:

Remember to use radians, not degrees, in arguments to trigonometric functions.


So I tried to use Radians instead and got -.000789 and that's wrong as well, lol.
 
  • #6
Well you know that the temperature of the crate should rise because it's absorbing heat energy. So any negative value you calculate is bound to be incorrect. You must still be having a calculator issue. Break the calculation into parts and let's see your intermediate results.
 
  • #7
gneill said:
Well you know that the temperature of the crate should rise because it's absorbing heat energy. So any negative value you calculate is bound to be incorrect. You must still be having a calculator issue. Break the calculation into parts and let's see your intermediate results.

True. Since it says to use radians, I did:

9.81*8.2*sin(37.6) and got -7.9597. Than I divided that by 3650 and got -.0021807. Than I subtracted (2.4)^2 from that and got -5.7622. Than I divided it by 2*3650 and got -.000789.
 
  • #8
Bigworldjust said:
True. Since it says to use radians, I did:

9.81*8.2*sin(37.6) and got -7.9597.
If you're going to use radians then the angle that you supply will have to be in radians! You've given the angle in degrees. Either convert the angle to radians before taking the sine, or change your calculator to accept degrees. You don't HAVE to use radians if you're given the angles in degrees to begin with; use whichever calculator mode suits the given problem.
Than I divided that by 3650 and got -.0021807. Than I subtracted (2.4)^2 from that and got -5.7622. Than I divided it by 2*3650 and got -.000789.

You should calculate the entire second term before you subtract it from the first one. The calculator will do the operations in the order entered; it does not know about your formula's layout on paper, it'll subtract 2.42 from the first term and then divide the result by 2 and then that result by 3650. That's not what you intended...
 
  • #9
gneill said:
If you're going to use radians then the angle that you supply will have to be in radians! You've given the angle in degrees. Either convert the angle to radians before taking the sine, or change your calculator to accept degrees. You don't HAVE to use radians if you're given the angles in degrees to begin with; use whichever calculator mode suits the given problem.You should calculate the entire second term before you subtract it from the first one. The calculator will do the operations in the order entered; it does not know about your formula's layout on paper, it'll subtract 2.42 from the first term and then divide the result by 2 and then that result by 3650. That's not what you intended...

Alright I did what you said and got .0134469-.000789 and got .01. That is still wrong tho, lmao. I don't get what I am doing wrong.
 
  • #10
Bigworldjust said:
Alright I did what you said and got .0134469-.000789 and got .01. That is still wrong tho, lmao. I don't get what I am doing wrong.

I think they'll want two significant figures. The leading zero after the decimal doesn't count.
 
  • #11
gneill said:
I think they'll want two significant figures. The leading zero after the decimal doesn't count.

Wow, you were right! That was pretty sneaky of them, lol. Thanks alot, I really appreciate it.
 
  • #12
Bigworldjust said:
Wow, you were right! That was pretty sneaky of them, lol. Thanks alot, I really appreciate it.

:biggrin: Glad to help.
 

1. How does sliding a crate affect its temperature?

Sliding a crate can increase its temperature due to the friction created between the crate and the surface it is sliding on. This friction converts mechanical energy into heat, resulting in a rise in temperature.

2. Is there a difference in temperature change between sliding a crate on different surfaces?

Yes, the temperature change can vary depending on the type of surface the crate is sliding on. Smooth surfaces with low friction, such as ice, will result in less temperature change compared to rough surfaces with higher friction, such as concrete.

3. How does the weight of the crate impact the temperature change during sliding?

The weight of the crate can affect the temperature change in two ways. Firstly, a heavier crate will create more friction with the surface, resulting in a higher temperature change. Secondly, a heavier crate may also require more force to slide, leading to more work being done and a greater increase in temperature.

4. Can the speed of sliding the crate affect its temperature change?

Yes, the speed at which the crate is slid can impact its temperature change. Sliding the crate at a higher speed will generate more friction and therefore result in a greater increase in temperature compared to sliding at a slower speed.

5. Are there any other factors that can affect the temperature change during sliding a crate?

Yes, there are other factors that can influence the temperature change during sliding, such as the material of the crate and the surface it is sliding on, the temperature and humidity of the environment, and the duration of sliding. These factors can all play a role in determining the final temperature change of the crate.

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