If 14.5 kJ of heat is added to 485 g of liquid water, how much will the temperature of the water increase? (Assume the specific heat capacity of water is 4.18 J/g·°C.)

• A. 12.0 °C
• B. 7.2 °C
• C. 3.5 °C
• D. 1.8 °C

Answer: (B)

Verified step by step guidance

1. Identify the known quantities: heat added \(q = 14.5\ \text{kJ}\), mass of water \(m = 485\ \text{g}\), and specific heat capacity of water \(c = 4.18\ \text{J/g} \cdot ^\circ\text{C}\).
2. Convert the heat added from kilojoules to joules because the specific heat capacity is in joules: \(q = 14.5\ \text{kJ} \times 1000\ \text{J/kJ}\).
3. Use the formula relating heat, mass, specific heat capacity, and temperature change: \(q = m \times c \times \Delta T\), where \(\Delta T\) is the temperature change.
4. Rearrange the formula to solve for the temperature change \(\Delta T\): \(\Delta T = \frac{q}{m \times c}\).
5. Substitute the known values into the equation and calculate \(\Delta T\) to find the increase in temperature.