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3 years ago

If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?

Chemistry

1 answer:

Y_Kistochka [10]3 years ago

5 0

Answer:

-26.125 kj

Explanation:

Given data:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 5.0°C - 30.0°C

ΔT = -25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m.c. ΔT

Q = 250.0 g × 4.18 j/g.°C × -25°C

Q = -26125 j

J to kJ

-26125 j ×1 kj /1000 j

-26.125 kj

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3 years ago

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Answer:

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Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ (1)

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ (2)

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

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$P^0_{A} = 490 torr$

Vapour pressure of cyclohexane at 50°C is 490torr

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$P^0_{B}=90torr$

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