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

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge