How many grams of c2h5oh must be burned to raise the temperature of 400.0 ml of water from 20.0 ∘c to 100.0 ∘c? (the specific heat of water is 1.00 cal/g⋅∘c or 4.184 j/(g⋅∘c)?
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
