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  • 15-02-2017
  • Chemistry
Answered

16.25 g of water at 54 C relaeases 402.7 J. What will be its final temp?

Verified Answer :

dexteright02
dexteright02 dexteright02
  • 15-02-2017

Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?

Solving:

Q = mcΔT
-402.7 = 16.254.184ΔT
-402.7 = 67.99*ΔT
\(\Delta\:T = \frac{-402.7}{67.99}\)
\(\boxed{\Delta\:T \approx -5.92\:^0C}\)

If: ΔT (T final - T initial) = ?
\(-5.92^0 = T_{final} - 54^0\)
\(T_{final} = 54^0 - 5.92^0\)
\(\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark\)

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