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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.

Respuesta :

abidemiokin

Answer:

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

snehashish65

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, [tex]L = 2.256 \times 10^{6} \;\rm J/kg[/tex].

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,

[tex]Q = m \times L[/tex]

Solving as,

[tex]Q = 0.025 \times (2.256 \times 10^{6})\\\\Q = 56400 \;\rm J[/tex]

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:

https://brainly.com/question/24203857

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