Answer
The amount of energy in these processes is equal to 9.6 kcal.
Work Step by Step
1. Identify each process.
a. Condensation of 15.0 g of steam at $100^{\circ}C$.
b. Cool the liquid to $0 ^oC$.
2. Find the conversion factors and formulas:
$1kcal = 1000 cal$
a. Heat of Vaporization (Water): $\frac{540cal}{1g}$ and $\frac{1g}{540cal}$.
b. Specific heat (SH) (Water): $\frac{1.00cal}{g^{\circ}C}$
$Heat = mass(g) \times \Delta T \times SH$
3. Calculate the energy in each process.
a. $15.0g \times \frac{540cal}{1g} = 8100cal$
b. $\Delta T = 100 ^o C - 0^oC = 100^o C$
$Heat = 15.0g \times 100 ^o C \times \frac{1.00cal}{g^{\circ}C} = 1500cal$
4. Sum these values to get the total heat, and convert it to kcal.
$(8100cal + 1500cal) \times \frac{1kcal}{1000cal} = 9.6kcal$
5. Adjust the number of significant figures.
Data with lowest number of SF's : 540 cal
