Answer
The total amount of energy is equal to 72.2 kJ.
Work Step by Step
1. Identify each process.
a. Melting of 24.0 g of ice at $0^{\circ}C$.
b. Warming the liquid to $100 ^oC$.
c. Boiling of liquid water at $100 ^o C$
2. Find the conversion factors and equations:
$1kJ = 1000J$
a. Heat of Fusion (Water): $\frac{334J}{1g}$ and $\frac{1g}{334J}$.
b. Specific heat (SH) (Water): $\frac{4.184J}{g^{\circ}C}$
$Heat = mass(g) \times \Delta T \times SH$
c. Heat of vaporizattion (water) : $\frac{2260J}{1g} and \frac{1g}{2260J}$
3. Calculate the energy in each process (Don't forget to adjust to the correct number of SF's.).
a. $24.0g \times \frac{334J}{1g} = 8016J = 8020J$
b. $\Delta T = 100 ^o C - 0^oC = 100^o C$
$Heat = 24.0g \times 100 ^o C \times \frac{4.184J}{g^{\circ}C} = 10041.6J = 1.00 \times 10^4J$
c. $24.0g \times \frac{2260J}{1g} = 54240J = 54200J$
4. Sum these values to get the total heat, and convert the value to kJ.
$(8020J + 1.00\times 10^4J + 54200J) \times \frac{1kJ}{1000J}= 72.22kJ = 72.2kJ$
