Question

1. how much energy would it take to heat 1.00 kg of ice at 0 °c to water at 15.0 °c?

a. 271 kj
b. 334 kj
c. 62.8 kj
d. 397 kj

Explanation:

Step1: Calculate ice melting energy

The heat required to melt ice is given by $Q_1 = m \cdot L_f$, where $m=1.00\ \text{kg}$ and $L_f=334\ \text{kJ/kg}$ (latent heat of fusion for water).
$Q_1 = 1.00\ \text{kg} \times 334\ \text{kJ/kg} = 334\ \text{kJ}$

Step2: Calculate water heating energy

The heat required to warm liquid water is $Q_2 = m \cdot c \cdot \Delta T$, where $c=4.186\ \text{kJ/(kg·°C)}$ and $\Delta T=15.0^\circ\text{C}-0^\circ\text{C}=15.0^\circ\text{C}$.
$Q_2 = 1.00\ \text{kg} \times 4.186\ \text{kJ/(kg·°C)} \times 15.0^\circ\text{C} = 62.79\ \text{kJ} \approx 62.8\ \text{kJ}$

Step3: Sum total energy

Total energy $Q_{total} = Q_1 + Q_2$
$Q_{total} = 334\ \text{kJ} + 62.8\ \text{kJ} = 396.8\ \text{kJ} \approx 397\ \text{kJ}$

Answer:

d. 397 kJ