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3 years ago

if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be

Chemistry

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

The final temperature of the water will be 328.81 K .

Explanation:

Using the equation, q = mcΔT

here, q = energy

m= mass

c= specific heat capacity

ΔT= change in temperature

Mass of water = 1kg (1000 g ) per liter

∴ 6.2 Liter of water = 6200 g

c of water ≈ 4.18 J /g/K

Now,

980000 = 6200*4.18*ΔT

ΔT = 37.81 K

∴ final temperature of the water = 291 + 37.81 = 328.81 K

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3 years ago

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2 years ago

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

leonid [27]

Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
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Solving:

Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
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$\Delta\:T = \frac{-402.7}{67.99}$
$\boxed{\Delta\:T \approx -5.92\:^0C}$

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

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