Two samples of water are mixed together.
the first sample has a mass of 0.50 kg and is at 0°c. the second sample has a mass of 1.5 kg and is at 100°c. what is the equilibrium temperature of the water, assume this is a closed system? (cwater = 4.18 kj/kg*°c.)
a.75°c
b.67°c
c.33°c
d.50°c

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x)            Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x)                          Remove the brackets.

0.5x = 150 - 1.5x                                  Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x             Combine like terms  

2x = 150                                               Divide by 2

x = 75

Answer

A

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14