Given that the specific heat capacity of water is 4.18 J/g·°C, how many grams of water can be heated from 20.0 °C to 75.0 °C using 12,500.0 joules of energy?

• A. 75.0 g
• B. 60.0 g
• C. 50.0 g
• D. 61.0 g

Answer: (D)

Verified step by step guidance

1. Identify the known values: specific heat capacity of water \(c = 4.18 \ \text{J/g} \cdot ^\circ\text{C}\), initial temperature \(T_i = 20.0 \ ^\circ\text{C}\), final temperature \(T_f = 75.0 \ ^\circ\text{C}\), and energy supplied \(q = 12,500.0 \ \text{J}\).
2. Calculate the temperature change \(\Delta T\) using the formula \(\Delta T = T_f - T_i\).
3. Use the heat transfer formula \(q = m \times c \times \Delta T\), where \(m\) is the mass of water in grams, to relate the given quantities.
4. Rearrange the formula to solve for the mass \(m\): \(m = \frac{q}{c \times \Delta T}\).
5. Substitute the known values of \(q\), \(c\), and \(\Delta T\) into the equation to find the mass of water that can be heated.