Athermally isolated system consists of a hot piece of aluminum and a cold piece of copper. the aluminum and the copper are in thermal contact. the specific heat of aluminum is more than double that of copper. which object experiences the greater amount of heat transfer during the time it takes the system to reach thermal equilibrium?

a) the aluminum experiences the greater magnitude of heat transfer.
b) the copper experiences the greater magnitude of heat transfer.
c) both experience the same magnitude of heat transfer.

Explanation:

Make up a question.

The only change is going to be c.

Suppose they aluminum starts our higher at 50oC

Suppose the copper starts out at 20oC

Suppose the mass of both are 25 grams.

Aluminum

m*2c * deltat

deltat = 50 - xc = 2*cm = 25

Copper

m*c*deltat

deltat = x - 20

m = 25

c = c

Now since the amount of heat is the same (this starts out on a heated slab of something).

m*2c * (50 - x) = m * c * x - 20  The m and the c are the same. Cancel them out.

2 * (50 - x) = (x - 20)    Remove the brackets.

100 - 2x = x - 20          Add 20 to both sides.

120 - 2x = x                  Add 2x to both sides.

120 = 3x                       Divide by 3

x = 40

What does this tell you?

It tells you that the temperature of the aluminum is only going to drop 10 degrees

The copper is going to gain 40 - 20 = 20  degrees.

The heat transfer is actually the same. It doesn't take as much heat to heat copper as it does aluminum. That's shown by the difference in how the temperature changes. One looses 10 degrees. The other gains 20. The transfer is the same because of the way the "c" operates.        

If there was no gravity then their would be no downword force on the football to have it return the the grownd, and if there was no friction (mainly air friction) then there would be no opposing force the stop the ball from moving through the air allowing the ball to travle ∞.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2