If 25.5 g of MgCl₂ are dissolved 500.0 g of water at 20.0 °C in an insulated container, a temperature change is observed. The ∆H of solution of MgCl₂ is -160.0 kJ/mol. Assuming that the specific ...

Question

If 25.5 g of MgCl₂ are dissolved 500.0 g of water at 20.0 °C in an insulated container, a temperature change is observed. The ∆H of solution of MgCl₂ is -160.0 kJ/mol. Assuming that the specific heat of the solution is 4.184 J/(g°C), and that no heat is gained or lost by the container, what will be the final temperature of the solution?

I keep getting an answer under 20 shouldnt it be over 20 because the MgCl2 is losing heat (because the change in heat is negative) so the solution should be gaining heat so it should be more than 20 because the initial temp was 20?

Answer 1

Yes. The final temperature should be >20º.

First, we'll convert 25.5 g MgCl2 to moles:

25.5 g MgCl2 x 1 mol / 95.2 g = 0.2679 moles

Now, find the heat generated upon dissolution of this many moles:

0.2679 mols x 160.0 kJ / mol = 42.864 kJ = 42864 J (I just like to work in J instead of kJ)

Now we use this amount of heat energy in the equation q = mC∆T:

q = heat = 42864 J

m = mass of water = 500.0 g (some add the 25.5 g but we won't at this level of chemistry)

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = ?

Solving for ∆T, we have...

∆T = 42864 J / (500 g)(4.184 J/g) = 20.5º

Add this to the original temperature of water to get the final temperature

Final temperature = 20.0º + 20.5º = 40.5º