What is the final temperature when 625 grams of water at 75.0 C loses 7.96 x 10 to the fourth power J ? The specific heat of water is 4.18 J/gC

The final temperature of the water was 44.5 °C

Let the final temperature be F. Therefore:

Heat lost = mass * specific heat of water * change in temperature

-7.96 * 10⁴ = 625 g * 4.18 J/gC * (F - 75)

F - 75 = -30.5

F - 75 + 75 = -30.5 + 75

F = 44.5 °C

Hence the final temperature of the water was 44.5 °C

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The final temperature : 44.5 °C

Heat can be formulated :

Q = m . c .Δt

Q = heat, J

m = mass, g

c = specific heat, J/gC

Δt= temperature

m= 625 g

Q=7.96 x 10⁴ J

c = 4.18 j/gC

t₁=75 C

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8