Calculate the change in enthalpy produced by dissolving 14.3 g of NaOH in 65.0 g of water if the temperature increases 49.1°C and the specific heat of water is 4.18 J/g°C
I believe you multiply 4.18, 65.0, and 49.1. Am I correct
Yes and if that's all the question wants you're ok. However, most questions like this want kJ/mol or kJ/gram (in this case mol or g of NaOH).
To calculate the change in enthalpy for dissolving NaOH in water, we need to consider two components: the heat absorbed by the water and the heat released by the NaOH.
First, let's calculate the heat absorbed by the water.
The formula to calculate heat is:
Q = m * c * ΔT
Where:
Q = heat (in Joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
Substituting the given values:
Q_water = 65.0 g * 4.18 J/g°C * 49.1°C
Now we can calculate the heat released by the NaOH.
The heat released by the NaOH is equal to the heat absorbed by the water, but with the opposite sign, as energy is conserved. So, the heat released by the NaOH is -Q_water.
Finally, the change in enthalpy is equal to the sum of the heat absorbed by the water and the heat released by the NaOH:
ΔH = Q_water + (-Q_water) = 0
Therefore, the change in enthalpy produced by dissolving 14.3 g of NaOH in 65.0 g of water is zero.