Question

A 6.50-g sample of liquid water at 25.0 ∘

C is heated by the addition of 145 J of energy. The final temperature of the water is ∘
C. The specific heat capacity of liquid water is 4.18 J/g ∘
C. −19.7 30.3 118 25.2 5.34

Answer 1

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.