Is the traditional method for calculating molar heat of dissolution accurate?

[PVnRT81]

Homework Statement

Suppose I have a solute of mass "x" grams. I dissolve this in "y" mL of water. The temperature of the water increases by "z" degrees Celsius.

I calculate the molar heat of dissolution of the solute by measuring, using q=m_watercΔt, the heat absorbed by water (in kJ), and then dividing by the moles of solute.

The above mentioned method is found in many reference textbooks as well as various sources across the web.

However, my science instructor insists that this method is wrong, and to calculate the heat transferred, you should include the mass of the solute as well.

Thus, when measuring the heat absorbed it should look like this:

q=m_{solute + solvent}cΔt

I find this quite unintuitive; I don't agree with my instructor on this matter. Am I right or wrong?

 

[Borek]

Wrong.

Imagine that after dissolution finished you are adding another, identical amount of heat, this time without dissolving anything. What mass would you use now to calculate Δt?

(To be exact your instructor is in a way also wrong, but (s)he is much closer to the correct answer. Their solution assumes c_p didn't change, which is not exactly true.)

 

[PVnRT81]

Wrong.

Imagine that after dissolution finished you are adding another, identical amount of heat, this time without dissolving anything. What mass would you use now to calculate Δt?

(To be exact your instructor is in a way also wrong, but (s)he is much closer to the correct answer. Their solution assumes c_p didn't change, which is not exactly true.)

I would use the mass that is gaining heat - which is the total mass in this case. But in the original scenario, the solute is losing heat, and the solvent is gaining heat... so, wouldn't it make more sense to use the mass of water only because the solute itself is not gaining heat?

 

[Borek]

Solute is not losing heat! Heat is produced by the dissolution, and it has to heat up both water and the solute. Otherwise after the dissolution you will have a hot solvent and cold solute - which is apparently not the case.

 

[PVnRT81]

Solute is not losing heat! Heat is produced by the dissolution, and it has to heat up both water and the solute. Otherwise after the dissolution you will have a hot solvent and cold solute - which is apparently not the case.

Ah! That clears it up... however, I am still curious as to why some websites/textbooks choose not to include the mass of the solute.

 

[Borek]

Sometimes it is negligible (much smaller than the mass of the solvent).