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Answer:
Explanation:
Hello,
In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:
Regards.
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.
