you had 25 grams of water, not 25kg. Rework your mc*deltaT with units.
Then divide by .0210
on the second reaction, how would you measure the heat given off? In a solution of water, it is easy, you use a balance and a thermometer.
Back again with another problem :)
How do you calculate the enthalpy change, in kJ/mol for the reaction of copper(II) sulfate?
0.0210 mol of copper sulfate was added to 25 cm3 of water. The temperature increased by 14°C. The specific heat capacity of water is 4.18 J k-1 g-1
What I did was get heat released with mcΔT, which was 25 x 4.18 x 14 = 1463
I then divided this by 0.0210 to get 69.7 kJ/mol.
Is this correct? Thanks for checking my answer!
Also, why can't we measure the value of ΔH for the reaction of turning anhydrous copper sulfate to hydrated copper sulfate directly? Why are we forced to make it into copper sulfate solution instead?
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