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If 2.0 g of water at 0.00°C is to be vaporized, how much heat must be added to it? The specific heat of water is 1.0 cal/g∙K, its heat of fusion is 80 cal/g, and its heat of vaporization is 539 cal/g.

Respuesta :

xero099

Answer:

[tex]Q_{vap} = 1438\,cal[/tex]

Explanation:

The heat needed to vaporize the water is:

[tex]Q_{vap} = (2.0\,g)\cdot[80\,\frac{cal}{g} +(1\,\frac{cal}{g\cdot K} )\cdot (100\,K)+539\,\frac{cal}{g} ][/tex]

[tex]Q_{vap} = 1438\,cal[/tex]

pstnonsonjoku

The heat required to vaporize 2 g of water at 0.00°C is 1078 cal.

We know that the heat required to vaporize the water is obtained from the relation;

H = mL

H = Heat required to vaporize a given mass of water

m = mass of water vaporized

L = Heat of vaporization of water

Vaporization is an endothermic process so heat is always added to the system in order to make it happen.

So;

H = 2.0 g ×  539 cal/g = 1078 cal

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