If 14.5 kJ of heat were added to 485 g of liquid water at an initial temperature of 25°C, what would be the final temperature of the water? (Assume the specific heat capacity of water is 4.18 J/g·°C.)

• A. 30.0°C
• B. 28.5°C
• C. 35.0°C
• D. 32.1°C

Answer: (D)

Verified step by step guidance

1. Convert the heat added from kilojoules to joules because the specific heat capacity is given in joules: Use the conversion \$1\ \text{kJ} = 1000\ \text{J}\(, so calculate \)q = 14.5\ \text{kJ} \times 1000$ to get joules.
2. Identify the known values: mass \(m = 485\ \text{g}\), specific heat capacity \(c = 4.18\ \text{J/g} \cdot ^\circ\text{C}\), initial temperature \(T_i = 25^\circ\text{C}\), and heat added \(q\) in joules from step 1.
3. Use the formula relating heat, mass, specific heat, and temperature change: \(q = m \times c \times \Delta T\), where \(\Delta T = T_f - T_i\) and \(T_f\) is the final temperature.
4. Rearrange the formula to solve for the final temperature \(T_f\): \(T_f = \frac{q}{m \times c} + T_i\).
5. Substitute the values of \(q\), \(m\), \(c\), and \(T_i\) into the equation from step 4 to calculate the final temperature \(T_f\).