How many kJ are required to completely vaporize a 25.0 gram sample of water at 100°C? general formula:

Chemistry

1 answer:

Reptile [31]2 years ago

5 0

Answer:

56500J

Explanation:

Given parameters:

Mass of water = 25g = 0.025kg

Temperature of the water = 100°C

Unknown:

Amount of heat required to vaporize the water = ?

Solution:

The amount of heat required to vaporize the water is a function of the mass of the water and the latent heat of vaporization of water.

H = mL

H is the amount of heat

m is the mass

L is the latent heat of vaporization of water = 22.6 x 10⁵J/Kg

H = 0.025 x 22.6 x 10⁵ = 56500J

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(b) 0.0000000778

(c) 0.000001

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Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as $5.0\times 10^3$

889.9 is written as $8.899\times 10^{-2}$

In this examples, 5000 and 889.9 are written in the standard notation and $5.0\times 10^3$ and $8.899\times 10^{-2}$ are written in the scientific notation.

(a) $1.52\times 10^{-2}$