What quantity of heat must be absorbed to convert 86.1 g of liquid to a vapor at its boiling point, -24.09 c? the heat of vaporization of ch3cl is 21.40
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To calculate the quantity of heat required to convert a given mass of liquid to vapor, we need to use the formula:
Q = m * ΔHvap
Where:
Q is the quantity of heat (in Joules),
m is the mass of the substance (in grams), and
ΔHvap is the heat of vaporization (in J/g).
In this case, we have:
m = 86.1 g
ΔHvap = 21.40 J/g
Substituting these values into the formula, we get:
Q = 86.1 g * 21.40 J/g
Now, multiply the mass (86.1 g) by the heat of vaporization (21.40 J/g) to find the quantity of heat:
Q = 1841.34 J
Therefore, the quantity of heat required to convert 86.1 g of liquid CH3Cl to a vapor at its boiling point is 1841.34 J.