Question

Show a numerical setup for calculating the quantity of heat in joules required to completely vaporize 102.3 grams of H2O(l) at 100°C and 1.0 atm.

Answer 1

Answer:

The heat required to vaporize 102.3 grams of H₂O(l) is 231198 J.

Explanation:

The heat required to vaporize 102.3 g of H₂O(l) can be calculated as follows:

$q = m*\Delta H_{v}$

Where:

q: is the heat  

ΔHv: is the heat of vaporization of water = 2260 J/g

m: is the mass = 102.3

$q = m*\Delta H_{v} = 102.3 g*2260 J/g = 231198 J$

Therefore, the heat required to vaporize 102.3 grams of H₂O(l) is 231198 J.

I hope it helps you!

Answer 2

The heat required to vaporize 102.3 grams of H₂O(l) is 231198 J.

The heat required to vaporize 102.3 g of H₂O(l) can be calculated as follows:

$q=m\times H_v$.....(1)

Here,  q is the heat  

$H_{v}$: is the heat of vaporization of water = 2260 J/g

m: is the mass = 102.3 g

Substitute the value in equation (1) as follows:-

$q=102.3\ g\times2260\ J/g\\\\=231198\ J$

Therefore, the heat required to vaporize 102.3 grams of H₂O(l) is 231198 J.

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