Answer
The amount of heat energy in these processes is equal to 30300J
Work Step by Step
1. Identify each process.
a. Melting of 50.0 g of ice at $0^{\circ}C$.
b. Warm the liquid to $65.0 ^oC$.
2. Find the conversion factors and formulas:
a. Heat of Fusion (Water): $\frac{334J}{1g}$ and $\frac{1g}{334J}$.
b. Specific heat (SH) (Water): $\frac{4.184J}{g^{\circ}C}$
$Heat = mass(g) \times \Delta T \times SH$
3. Calculate the energy in each process.
a. $50.0g \times \frac{334J}{1g} = 16700J$
b. $\Delta T = 65.0 ^o C - 0^oC = 65.0^o C$
$Heat = 50.0g \times 65.0 ^o C \times \frac{4.184J}{g^{\circ}C} = 13598J$
4. Sum these values to get the total heat.
$16700J + 13598J = 30298J$
5. Adjust the number of significant figures.
Data with lowest number of SF's : 50.0 g
