|
GR9677 #14
|
|
|
|
|
Alternate Solutions |
| There are no Alternate Solutions for this problem. Be the first to post one! |
|
|
Comments |
Phys4
2010-08-02 15:44:18 | Curious, the version of the exam I have (GR9677 obtained from the Ohio State website) defines the specific heat in units of Kelvin and not Celsius. To compute this problem straight through, I would believe that you would need to convert from Celsius to Kelvin. In this case, since you don't, am I safe to assume that the original exam had an error? |  | wittensdog
2009-10-16 14:18:22 | Just a little side note, which may or may not have been implicitly assumed here in the solution (I'm not sure which way was used to attain the value of fifty degrees). Whenever you have two identical bodies, the resulting temperature is the average of the initial ones. It would be very nice of ETS if all of their problems had nice little features like this, but I certainly wouldn't count on it! In general you would need to use,
m1*c1*(T1 - Tf) + m2*c2*(T2 - Tf) = 0
and then solve for Tf. Then you can plug Tf back into one of the two terms above to find the heat transfer. | | Ning Bao
2008-01-30 10:19:22 | Dimensional Analysis:
They give you a constant that has units A/BC. Average the two temperatures to get something of units C. B is given. Multiply for the win.
JMOAN
2008-10-30 20:33:54 |
Just a warning on Ning Bao's comment:
This assumes that there aren't other attributes to the solution (such as multiplication by a scaler), which happens to be true for this problem, but is not always the case. In general, we can only use simple dimensional analysis to eliminate answers that have the wrong dimensions (this problem's solutions all have the same dimensionality). Otherwise, you might think that kinetic energy, which has units of  , is just calculated by multiplying mass by the square of velocity, when in fact you must multiply this result by 0.5: 
|
| |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
