If 36.5 kJ of heat is added to 275 g of water, by how many degrees Celsius 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. 31.0 °C
• C. 3.1 °C
• D. 48.0 °C

Answer: (A)

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 factor: 1 kJ = 1000 J. So, multiply 36.5 kJ by 1000 to get the heat in joules.
2. Recall the formula relating heat added, mass, specific heat capacity, and temperature change: \(q = m \times c \times \Delta T\), where \(q\) is heat in joules, \(m\) is mass in grams, \(c\) is specific heat capacity in J/g·°C, and \(\Delta T\) is the temperature change in °C.
3. Rearrange the formula to solve for the temperature change \(\Delta T\): \(\Delta T = \frac{q}{m \times c}\).
4. Substitute the known values into the rearranged formula: use the heat in joules from step 1, the mass of water (275 g), and the specific heat capacity of water (4.18 J/g·°C).
5. Calculate \(\Delta T\) by performing the division to find the temperature increase in degrees Celsius.