Answer to Question #96417 in General Chemistry for Delaney Kilfoil

Question

How many kilojuoles of heat are absorbed when 1.00 L of water is heated from 18\deg C to 85\deg C?

Answer 1

Q = m х C х ΔT

m = mass of substance in grams;

C = specific heat capacity of substance in J/g°C

ΔT = temperature difference.

C = 4.18 J/g°C

ΔT = (85 - 18) °C = 67 °C

Volume of water = 1 L, density of water = 1 g/mL

mass of water = 1000 g

Q = 1000 g x 4.18 J/g°C x 67 °C = 280060 J = 280.06 kJ