How much energy is required to melt 13.00 kg of ice at 273 K? The specific latent heat of fusion of ice is 3.36 x 10^5 J kg-1.
A. 1.19 x 10^9 J

B. 2.58 x 10^4 J

C. 4.37 x 10^6 J

D. 1174.83 J

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cristisp93 cristisp93 · Answer 1 · 2019-07-23T12:28:59+02:00

You use the information given by the question about latent heat of fusion of ice and construct the following reasoning:

if you need 3.36 × 10⁵ J to melt 1 kg of ice

then you need X J to melt 13 kg of ice

X = (13 × 3.36 × 10⁵)/1 = 43.68 × 10⁵ J = 4.368 × 10⁶ J ≈ 4.37 × 10⁶ J

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Alleei Alleei · Answer 2 · 2019-08-13T09:52:51+02:00

Answer : The amount of heat required is, (C) [tex]4.37\times 10^6J[/tex]

Explanation :

Formula used :

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

where,

q = heat required = ?

L = specific latent heat of fusion of ice = [tex]3.36\times 10^5J/kg[/tex]

m = mass of ice = 13.00 kg

Now put all the given values in the above formula, we get:

[tex]q=(3.36\times 10^5J/kg)\times (13.00kg)[/tex]

[tex]q=4.37\times 10^6J[/tex]

Therefore, the amount of heat required is, [tex]4.37\times 10^6J[/tex]