Temperature change with sliding a crate

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Homework Help Overview

The discussion revolves around a physics problem involving a crate of fruit sliding down a ramp. The crate has a specified mass and specific heat capacity, and the problem asks for the temperature change of the crate as it absorbs heat equal to the work done by friction. Participants are exploring the calculations related to temperature change based on the given parameters.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for temperature change and share their calculations, questioning the accuracy of their results. There is a focus on the correct use of degrees versus radians in trigonometric functions, as well as the significance of significant figures in the final answer.

Discussion Status

The discussion is ongoing, with participants providing guidance to each other regarding calculation methods and potential errors. Some participants have suggested breaking down calculations into parts to identify mistakes, while others have pointed out the importance of calculator settings for angle measurements.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific requirements for significant figures and the use of degrees or radians in calculations. There is also a mention of a previous question that provided additional context for the speed of the crate.

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

:biggrin: Glad to help.
 

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