Question

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)

Answer 1

T_f = 44.6 ° C

Step-by-step explanation:

We can solve this exercise using the calorimetry relations

Q = m c_e ΔT

ΔT =
`(Q)/(m \ c_e)`

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