# Temperature change with sliding a crate

## 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!

gneill
Mentor

## 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?

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

gneill
Mentor
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.

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.

gneill
Mentor
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.

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.

gneill
Mentor
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...

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.

gneill
Mentor
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.

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.

gneill
Mentor
Wow, you were right! That was pretty sneaky of them, lol. Thanks alot, I really appreciate it. Glad to help.