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Section 1
What is the energy required to evaporate 2 Kg of liquid potassium? The latent heat of evaporation of
potassium is 1970 KJ/kg

Respuesta :

Eduard22sly

The energy required to evaporate 2 Kg of liquid potassium is 3940 J

Data obtained from the question

From the question given above, the following data were obtained:

  • Mass of potassium (m) = 2 kilogram
  • Latent heat of evaporation of potassium (L) = 1970 KJ/kg
  • Heat energy (Q) =?

How to determine the energy required

The energy required to evaporate the the liquid potassium, can be obtain as follow:

Q = mL

Q = 2 × 1970

Q = 3940 J

Therefore, the heat energy required to evaporate the liquid potassium is 3940 J

Learn more about heat transfer:

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