Final answer:
The final temperature of the water sample is approximately 27.98 °C.
Step-by-step explanation:
The heat absorbed by the water can be determined using the equation:
q = mcΔT
where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, the mass of the water is 10.0 grams, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is unknown.
Substituting the values into the equation:
209 J = (10.0 g) x (4.18 J/g°C) x ΔT
Solving for ΔT, we get:
ΔT = 209 J / ((10.0 g) x (4.18 J/g°C))
ΔT ≈ 4.98 °C
Therefore, the final temperature of the water sample is approximately 23.0 °C + 4.98 °C = 27.98 °C.