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

Answer: 4.5 grams