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A heater marked 50W will evaporate 0.005kg of boiling water in 50 seconds
What is the specific latent heat of vaporization of water in J/kg?

Sagot :

MathPhys

Answer:

5.0×10⁵ J/kg

Explanation:

The specific latent heat of vaporization (L) is the amount of energy (E) per mass (m) needed for a liquid at its boiling point to change states into a vapor. The energy (E) absorbed by the boiling water is equal to the power (P) of the heater times the amount of time (t).

The amount of energy is:

E = Pt

E = (50 W) (50 s)

E = 2500 J

The specific latent heat of vaporization is:

L = E/m

L = (2500 J) / (0.005 kg)

L = 500,000 J/kg

L = 5.0×10⁵ J/kg

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