Question

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 3.40 kg of water to go from -19.0 °C to 64.0 °C.

Melting point	0.0	°C
Boiling point	100.0	°C
?Hfus	6.01	kJ/mol
?Hvap	40.67	kJ/mol
cp (s)	37.1	J/mol ·°C
cp (l)	75.3	J/mol ·°C
cp (g)	33.6	J/mol ·°C

Answer 1

3.40 kg of water = 3400 gm water = mass / molar mass = 3400 / 18 = 188.9 mole.

amount of energy needed = n1 * Cp(s) * dT1 (- 19 to o oC) + n1 * delta Hfus + n1 * Cp(l) * dT2 (from 0 to 64 oC)

amount of energy needed = 188.9 * 37.1 * (0 + 19) + 188.9 * (6.01 * 1000) + 188.9 * 75.3 * (64 - 0)

amount of energy needed = 133155.6 + 1135289 + 910346.9

amount of energy needed = 2178791.5 J = 2178.8 KJ (answer)