Question

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

Answer 1

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 = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-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) = ?
$-5.92^0 = T_{final} - 54^0$
$T_{final} = 54^0 - 5.92^0$
$\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark$