[(1.7x10^6) / (2.62x10^5)]+7.18
1 answer:
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<u>Answer:</u> The amount of energy absorbed by water is 5390 Calories
<u>Explanation:</u>
To calculate the amount of heat absorbed at normal boiling point, we use the equation:
where,
q = amount of heat absorbed = ?
m = mass of water = 10 grams
= latent heat of vaporization = 539 Cal/g
Putting values in above equation, we get:
Hence, the amount of energy absorbed by water is 5390 Calories