A155.0 g piece of copper at 168 oc is dropped into 250.0 g of water at 20.9 oc. (the specific heat of copper is 0.385 j/goc.) calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

Note that
The specific heat of water is 4.184 J/(g -C)

Let x = the final equilibrium temperature.

Heat loss by the copper is
(155 g)*(0.385 J/(g -C))*(168 - x C) = 59.675(168 -x) J

Heat gain by the water is
(250 g)*(4.184 J/(g-C))*(x - 20.9 C) = 1046(x - 20.9) J

Because there is no heat loss to the surroundings, therefore
59.675(168 - x) = 1046(x - 20.9)
168 - x = 17.5283(x - 20.9)
18.5283x = 534.3415
x = 28.84 °C

  28.8 °°C  nearest tenth)

10 is the . . .

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.