Question

what is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point

7.97
53.9
5390
7970

Answer 1

Answer: The amount of energy absorbed by water is 5390 Calories

Explanation:

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

$q=m\times L_{vap}$

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

$L_{vap}$ = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

$q=10g\times 539Cal/g=5390Cal$

Hence, the amount of energy absorbed by water is 5390 Calories

Answer 2

C. 5390

The amount of energy absorbed by water is 5390 Calories

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

$q=m*L_{vap}$

where,

q = amount of heat absorbed

m = mass of water  

$L_{vap}$ = latent heat of vaporization

Given:

m= 10 grams

It is known that the latent heat of vaporization of water is ≈540 cal/g

∴ $L_{vap}$ = 539  Cal/g

Putting values in above equation, we get:

$q=10 g*539 Cal/g\\\\q=5390 Cal$

Thus, the amount of energy absorbed by water is 5390 Calories

Therefore, option C is correct.

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