How many kilojoules of heat are absorbed when 2.00 L of water is heated from 19*C to 80*C? (Show all work including the equation you use to solve)

The answer to your question is 510.7 kJ

Explanation:

Data

Q = ?

Volume = 2 L

Temperature 1 = T1 = 19°C

Temperature 2 = T2 = 80°C

Specific heat = 4.186 J/kg°C

Density of water = 1 g/ml

Equation

              Q = mC(T2 - T1)

Process

1.- Calculate the mass of water

mass = density x volume

mass = 1 x 2000

mass = 2000 g

2.- Substitute values in the Heat formula

              Q = (2000)(4.186)(80 - 19)

-Simplification

               Q = (2000)(4.186)(61)

-Result

              Q = 510692 J  or 510.7 kJ