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While water is boiling, it absorbs 2257 kJ of heat per kilogram of water. This takes place at a
constant temperature of 100°C. What is the entropy change of 10 kg of water while it boils? (Be
careful of your energy units).

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Cricetus

Answer:

The correct answer is "6.015×10⁴ J/K".

Explanation:

The given values are:

Mass of water,

= 10 kg

Latent heat of boiling,

= 2257 kJ

Constant temperature,

= 100°C

The entropy change will be:

= [tex]\frac{mass\times latent \ heat \ of \ boiling}{temperature}[/tex]

On substituting the values, we get

= [tex]\frac{10\times 2257\times 1000}{(273+100)K}[/tex]

= [tex]\frac{22,570,000}{373K}[/tex]

= [tex]6.051\times 10^4 \ J/K[/tex]

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