From tables, the heat of combustion of ethanol (C₂H₅OH) is 29.7 kJ/g.
Given:
V = 400 mL of water
ΔT = 100 - 20 = 80°C, temperature rise
c = 1.00 cal/(g-°C), specific heat of water = 4.184 J/(g-°C)
Because the density of water is approximately 1.0 g/mL. the mass of water is
m = 400 g.
Let x = grams of ethanol that should be burned to make the water rise
by 80 °C.
Then
(29.7 x 10³ J/g)*(x g) = (400 g)*(4.184 J/g-°C))*(80 °C)
29.7 x 10³x = 1.339 x 10⁵
x = 4.5 g
Answer: 4.5 grams
