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vichka [17]
3 years ago
14

If you start with 163 g of water at 29◦C, how much heat must you add to convert all the liquid into vapor at 100◦C? Assume no he

at is lost to the surroundings.
Answer in units of kJ.
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

48.37514 kj

Explanation:

Given data:

Mass of water = 163 g

Initial temperature = 29°C

Final temperature = 100°C

Heat added = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C

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 = Final temperature - initial temperature

ΔT =  100°C - 29°C

ΔT =  71°C

Q = 163 g × 4.18 j/g.°C × 71°C

Q = 48375.14 j

Joule to Kj conversion:

48375.14 /1000 = 48.37514 kj

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Total mass of sample = 100 g

Mass of carbon = 27.2 g

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To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

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  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

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For Oxygen  = \frac{3.57}{3.57}=1

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C_1:C_2=\frac{1}{2}:1=1:2

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