How much heat Q2Q2 is transferred to the skin by 25.0 gg of steam onto the skin? The heat of vaporization for steam is L=2.256×106J/kgL=2.256×106J/kg.

The amount of heat transfer from the steam is of 56400 J.

Given data:

The mass of steam is, m = 25.0 g = 0.025 kg.

The latent heat of vaporization of steam is, .

When heat is to be given to a unit mass of substance to converts it from a liquid phase to vapor phase. Then it is known as Latent heat of vaporization. The amount of heat is expressed as,

Solving as,

Thus, we can conclude that the amount of heat transfer from the steam is of 56400 J.

Learn more about the Latent heat of vaporization here:

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56400Joules

Explanation:

The quantity of heat required is expressed as;

Q = mL

m is the mass = 25g = 0.025kg

L is the latent heat of vaporization for steam = 2.256×10^6J/kg

Substitute into the formula as shown;

Q = 0.025×2.256×10^6

Q = 56400Joules

Hence the quantity of hear required is 56400Joules

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