Answer
The total amount of heat energy is equal to 50.kcal.
Work Step by Step
1. Identify each process.
a. Melting of 525 g of ice at $0^{\circ}C$.
b. Warming the liquid to $15.0 ^oC$.
2. Find the conversion factors and equations:
$1kcal = 1000cal$
a. Heat of Fusion (Water): $\frac{80.cal}{1g}$ and $\frac{1g}{80.cal}$.
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 (Don't forget to adjust the values to the correct number of SF's.).
a. $525g \times \frac{80.cal}{1g} = 42000cal$
b. $\Delta T = 15.0 ^o C - 0^oC = 15.0^o C$
$Heat = 525g \times 15.0 ^o C \times \frac{1.00cal}{g^{\circ}C} = 7875 cal = 7880 cal$
4. Sum these values to get the total heat, and convert the value to kcal.
$(42000cal + 7880cal) \times \frac{1kcal}{1000cal} = 49.88 kcal = 50.kcal$
