Design and perform an experiment that will determine the dHrxn for the dissolution of H2SO4 in water.

Allowing for only PV work at constant pressure dH = qp; so, we are looking for the amount of heat transferred during the reaction.

How can we determine the heat of a reaction?

When we discussed the transfer of heat to water we measured the temperature change of the water.

Can we measure the temperature change of a reaction? No, we can only measure the temperature of objects. A reaction is not an object; it is a process.

The key to calorimetry is that all of the heat released or absorbed by the thing being studied

in this case the dissolution of H2SO4  

is absorbed by the calorimeter and its contents, which are usually water.

That is,

-qthing = qcalorimeter


So, to find dHrxn we construct a calorimeter, and perform the reaction in the calorimeter. All the heat released by the reaction is absorbed by the calorimeter. Since we can measure the change in temperature for the calorimeter we can determine the amount of heat the calorimeter absorbed.

-qrxn = qcalorimeter

-qrxn = C dT

Since the calorimeter is made of water, a stirrer, a thermometer, and the container, we break the qcalorimeter into two terms.

-qrxn = qcal + qwater

This way we do not always have to add exactly
the same amount of water to the calorimeter.

-qrxn = Ccal dT + m s dT


We can measure the mass of the water which is used, we can measure the temperature change, and we know the specific heat of water. However, we do not know Ccal!

So, before we can do our experiment we must determine Ccal.

How would you determine Ccal? In the equation above qrxn and Ccal are the only unknown quantities. If qrxn were known, then we could solve for Ccal. We can determine Ccal by adding a known amount of heat and measuring the temperature change.

The easiest way to add a known amount of heat is to add hot water to a calorimeter filled with cold water.

-qHW = qcalorimter

-qrxn = qcal + qwater

-m s dTHW = Ccal dTCW + m s dTCW

dT for the calorimeter is the
same as
dT for the cold water

Add 50.0 g of water which is at 100.0 °C to our calorimeter which contains 50.0 g of water at 23.0 °C. The final temperature of the calorimeter is 59.0 °C.

-(50.0 g)(4.184 J g-1 °C-1)(59 °C - 100 °C) =

Ccal(59 °C - 23 °C) + (50.0 g)(4.184 J g-1 °C-1)(59 °C - 23 °C)


8577.2 J = Ccal(36.0) + 7531.2 J

1046 J = Ccal (36.0)

Ccal = 29.1 J K-1

Now that we know the heat capacity of our calorimeter we can use our calorimeter to determine the amount of heat a reaction releases.


3.20 g of H2SO4 were dissolved in 100.0 g of water, and the temperature went from 23 °C to 29.8 °C. Determine dH for the dissolution of H2SO4 in water.

-qrxn = qcal + qwater

-qrxn = Ccal(29.8-23) + m s (29.8-23)

Since the H2SO4 solution is actually the thing that absorbs the heat, we have to use the mass of the solution. Here is where we are going to make a small assumption. The specific heat capacity of the solution is the same as the specific heat capacity of water.

-qrxn = (29.1)(29.8-23) + (103.2)(4.184)(29.8-23)

-qrxn = 3134.0 J

qrxn = - 3134.0 J

Since dHrxn is expressed in kJ/mol we are not done yet.

How many moles of acid were dissolved?


summary of calorimetry