# Heat transfer and friction

1. May 7, 2005

### WY

hey i'm wondering if someone can help un-stuck me on this question - I'm not sure if what i'm diong is right ><

A crate of fruit with a mass of 31.0kg and a specific heat capacity of 3800 J/kg K slides down a ramp inclined at an angle of 38.4 below the horizontal. The ramp has a length of 9.00 m.

I got the first part of the question which was:
If the crate was at rest at the top of the incline and has a speed of 2.20 m/s at the bottom, how much work W_f was done on the crate by friction?

and then the next part is:
f an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change DeltaT?

so i went ahead and did it using Q=mcdeltaT with Q = W_f and m= 31 and c= 3800 to work out delta T or should m = mgsin38.4? I can't seem to figure what mass I need to use!

2. May 7, 2005

### OlderDan

All of the fruit is being warmed up by absorbing the heat generated, so m is what you need. m = mgsin38.4 does not make sense for a couple of reasons. First, it is dimensionally incorrect. The left side is mass and the right side is force. Second, mass is a scalar quantitiy; it does not have direction associated with it, so it does not have components in different directions the way vectors do. There may be reasons to divide mass into parts in some problems (not this one), but that division will not involve direction angles.

3. May 7, 2005

### WY

Hey that does make sense thanks!
But when I tried doing that -since i got W_f=-1630
I would calculate that -1630 = 31x3800xdeltaT
which would be -0.0138 - but this is wrong - have i missed out using a formula? or substituted in the wrong numbers?

4. May 8, 2005

### Galileo

Apart from the minus sign (heat is added to the crate so $\Delta Q]$ is positive) it looks allright to me. Don't know why it would be wrong...