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kobusy [5.1K]
3 years ago
8

Water (2350 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the ini

tial temperature of the water? Express your answer with the appropriate units.
Chemistry
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

40.7062 °C  

Explanation:

Let the initial temperature = x °C

Boiling temperature of water = 100 °C

Using,

Q = m C ×ΔT

Where,  

Q is the heat absorbed in the temperature change from x °C to 100 °C.

C gas is the specific heat of the water = 4.184 J/g  °C

m is the mass of water

ΔT = (100 - x) °C  

Given,

Mass = 2350 g

Q = 5.83 × 10⁵ J

Applying the values as:

Q = m C ×ΔT

5.83 × 10⁵ = 2350 × 4.184 × (100 - x)

<u>x, Initial temperature = 40.7062 °C  </u>

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Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,
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Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

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The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

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