PLEASE HELP ME WITH THIS ONE QUESTION What is the final temperature when 625 grams of water at 75.0° C loses 7.96 x 10^4 J? (cwater = 4186 J/kg°C)

T_f = 44.6 ° C

Explanation:

We can solve this exercise using the calorimetry relations

          Q = m c_e ΔT

          ΔT =

let's calculate

          ΔT = -7.96 10⁴ /(0.625 4186)

          ΔT = -3.04 10¹

The negative sign indicates that the temperature decreases

the temperature variation is

          T_f - T₀ = ΔT

          T_f = T₀ + ΔT

          T_f = 75.0 - 30.4

          T_f = 44.6 ° C

the first, second, and fifth options

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