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if 145kl of energy is added to water, what mass of water can be heated from 35C to 100C then vaporized at 100C

Sagot :

hamzaahmeds

Answer:

m = 0.057 kg = 57 g

Explanation:

Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water

[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]

where,

E = Energy added to water = 145 KJ

m = mass of water = ?

C = specific heat capacity of water = 4.2 KJ/kg.°C

ΔT = change in temperature = 100°C - 35°C = 65°C

H = Latent heat of vaporization of water = 2260 KJ/kg

Therefore,

[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]

m = 0.057 kg = 57 g

Lanuel

The mass of water that can be heated is equal to 0.527 kilograms.

Given the following data:

  • Quantity of energy = 145 kJ = 145,000 Joules.
  • Initial temperature = 35.0°C
  • Final temperature = 100.0°C

Scientific data:

  • Specific heat capacity of water = 4200 J/kg°C
  • Latent heat of vaporization of water = 2260 KJ/kg

To calculate the mass of water that can be heated:

The quantity of energy and heat.

Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.

Mathematically, this is given by this expression:

[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]

Making m the subject of formula, we have:

[tex]m=\frac{E}{c\theta + H}[/tex]

Substituting the parameters into the formula, we have;

[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]

Mass, m = 0.527 kilograms.

Read more on quantity of energy here: https://brainly.com/question/13439286

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