The enthalpy change for a chemical reaction is ΔH_{1} at temperature T_{1} and ΔH_{2} at temperature T_{2}. In science we always show a change in a variable as final minus original. So the change in temperature is T_{2} – T_{1} which leads to a change in the reaction enthalpy of ΔH_{2} — ΔH_{1}.

**ΔH _{2} — ΔH_{1} = C_{p} (T_{2}—T_{1}) **

The constant C_{p} shows how sensitive the reaction enthalpy is to a change in temperature and C_{p}, is called the heat capacity at constant pressure. Rearrange the above equation to get T_{1} as the subject of the equation.

**Solution**

The lower temperature T_{1} is inside the bracket and so we need to multiply out the bracket, rearrange the equation to get C_{p} T_{1} as the subject and finally divide throughout by C_{p}.

**ΔH _{2} — ΔH_{1} = C_{p} T_{2} — C_{p} T_{1} **

**C _{p} T_{1} = C_{p} T_{2} – ΔH_{2} + ΔH_{1}**

**T _{1} = T_{2} – ((ΔH_{2}/ C_{p}) + (ΔH_{1}+ C_{p}))**

This is perfectly correct but you may also have taken it one step further to the alternative, neater form.

**T _{1} = T_{2} – ((ΔH_{2} — ΔH_{1})/ C_{p})**