Final answer:
The specific heat capacity of water is 4.184 J/g °C. Given a 6.50-g sample of water initially at 25.0 °C, the final temperature of the water after adding 145 J of energy is 30.3 °C.
Step-by-step explanation:
The specific heat capacity of water is 4.184 J/g °C. In this problem, we have a 6.50-g sample of water that is initially at 25.0 °C. By adding 145 J of energy, we need to find the final temperature of the water. We can use the formula:
Heat energy = mass × specific heat capacity × change in temperature
Plugging in the given values, we have:
145 J = 6.50 g × 4.184 J/g °C × (final temperature - 25.0 °C)
Solving for the final temperature, we find that the water will reach a temperature of 30.3 °C.