Calculate the chage in enthalpy produced by dissolving 14.3 g 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.
So this is just q=mcΔt
Do you add 14.3 and 65.0 to use for m? Also would the sign for enthalpy be negative because heat is lost by the reaction and gained by the surroundings, which increased in temperature?
No, don't add the masses.
enthlapy is negative when heat is given off. In this case, it is +
To calculate the change in enthalpy (ΔH) of this process, we can use the formula:
ΔH = q = mcΔt
In this case, we are dissolving 14.3 g of NaOH in 65.0 g of water. The specific heat capacity (c) of water is given as 4.18 J/g°C, and the temperature change (Δt) is 49.1°C.
First, let's calculate the heat (q) transferred in this process:
q = mcΔt
q = (65.0 g) * (4.18 J/g°C) * (49.1°C)
q = 13544.33 J
The value of q is positive because energy is gained by the water.
Now, to calculate the change in enthalpy (ΔH), we need to consider the sign convention. Since heat is lost by the reaction (NaOH dissolving) and gained by the surroundings (water), the sign for ΔH is negative.
Therefore, the change in enthalpy (ΔH) is:
ΔH = -13544.33 J
Note: The negative sign indicates an exothermic process in which heat is released.