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If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporization of water is 22.6*10^5 J/kg.

a. 429 g
b. 258 g
c. 340 g
d. 600 g

Answer :

Jattoabdulhayyu

Answer:A

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mickymike92

Based on the heat of vaporization of water, the mass of water left after 700 kJ of heat is added to 700 g of water at 70 degrees C is 429 g.

What mass of water is evaporated?

To determine the mass of water evaporated, we first determine the heat required to raise water at 70° C to water at 100°C.

Heat required = mass * specific heat capacity of water * temperature change.

Heat required = 700 * 4.18 * 30

Heat required = 87780 J

Heat left to evaporate water = 700000 J - 87780 = 612220 J

Heat required to vaporize water = mass x latent heat of vaporization of water

Mass of water vaporized = Heat/ latent heat of vaporization of water

Mass of water vaporized =  612220 J/ 22.6*10^5 J/kg.

Mass of water vaporized = 0.271 kg = 271 g

Mass of water left = 700 - 271 g

Mass of water left = 429 g

Therefore, the mass of water left after 700 kJ of heat is added to 700 g of water at 70 degrees C is 429 g.

Learn more about heat of vaporization at: https://brainly.com/question/486624

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