Answer:
The amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C is 141.74 kJ
Explanation:
The latent heat of vaporization is the heat required to transform a giving mass of a liquid completely to vapor at its boiling point thus
The mass of water = 62.6 g
The latent heat of vaporization of water = 40.79 kJ/mol
The quantity of heat = Q = LH[tex]_{Vaporization}[/tex]·m where Q = required heat energy,
However the molar mass of water = 18.01528 g/mol
Therefore 62.6 g of water contains (62.6 g)/(18.01528 g/mol) = 3.47 moles
∴ Q = 3.47 moles × 40.79 kJ/mol = 141.74 kJ
The required amount of heat to convert 62.6 g of water to steam at 100 °C is 141.74 kJ
The amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C is 141.86kJ
The formula for calculating the required amount of heat is expressed according to the equation;
[tex]Q=\frac{m}{M}\times \triangle H_{vap}[/tex]
Given the following parameters
Substitute the given parameters into the formula to have:
[tex]Q=\frac{62.6}{18} \times 40.79\\Q= 141.86Joules[/tex]
Hence the amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C is 141.86kJ
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