How much heat is needed to completely boil 85.0 grams of water at 100.0 degrees Celsius?

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Answer 1 · 2018-05-06T14:26:39

$191.87$ $191.87$ kilojoules

We use the latent heat formula, which states that,

$q = m L$ $q=mL$

where:

$m$ $m$ is the mass of the substance

$L$ $L$ is the latent heat of the substance, in this case it is the latent heat of vaporization

$L_{f}$ $L_f$ of water is $40.65 \ "kJ/mol"$ . So, we might as well convert the mass of water into moles for easier calculations.

Here, we got:

$(85color(red)cancelcolor(black) "g")/(18.01528color(red)cancelcolor(black)"g""/mol")~~4.72 \ "mol"$

So, the energy needed is:

$q=4.72color(red)cancelcolor(black)"mol"*(40.65 \ "kJ")/(color(red)cancelcolor(black)"mol")$

$=191.87 \ "kJ"$