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How many kilojoules of energy are required to heat 50.0 g of ethyl alcohol from 60.0 °C to 78.4 °C and vaporize it? The specific heat of ethyl alcohol is 2.138 J/g°C; heat of vaporization is 853 J/g.

Sagot :

Answer:

44.62 kJ

Explanation:

Firstly, we calculate the energy needed to heat the liquid (ethyl alcohol) by using the formula:

Q = m × c × ∆T

Where;

Q = Amount of heat (J)

m = mass (g)

c = specific heat of ethyl alcohol = 2.138 J/g°C

∆T = change in temperature (°C)

According to the information given in this question;

Q = ?, m = 50.0g, ∆T = (78.4°C - 60°C) = 18.4°C

Therefore, using Q = mc∆T

Q = 50 × 2.138 × 18.4

Q (amount of energy needed to heat ethyl alcohol) = 1966.96 J

Next, we calculate and add the amount of heat needed to vaporize by using the formula;

How many kilojoules of energy are required to heat 50.0 g of ethyl alcohol from 60.0 °C to 78.4 °C and vaporize it? The specific heat of ethyl alcohol is heat of vaporization is 853 J/g.