Chemistry / Thermochemistry / Principles and Exploration
Calorimetry — principles and investigation
When estimating the heat released from the reaction as the temperature rise of the solution, the heat received by the calorimeter and the heat lost to the surroundings must also be considered. Change the reaction moles, solution mass, calorimeter heat capacity, and cooling loss. This activity compares the highest observed temperature with the value corrected by the cooling curve.
How does it work?
Heat loss reduces the observed temperature rise. Cooling curve extrapolation approximates the loss, but outliers and residuals remain.
−nΔH = (m·c + Ccal)ΔT
Try this investigation
- Starting at 20 °C, heat is generated steadily for 20 s and observed until 120 s. When dissipation is turned on, heat is released to the surroundings during and after heating.
- Record the same response as no loss or presence of loss and compare the observed highest temperature and extrapolation correction temperature with the outlier value.
- Compare the temperature change by changing only the number of moles or the calorimetric heat capacity.
Interpreting your results
What is the combined heat capacity of the solution and calorimeter?
The solution and container receive the heat of reaction together. qSolution=−nΔH=(mc+Ccal)ΔT and the sign of q is opposite to that of the reaction.
Assumptions and limitations
It's a hypothetical exothermic reaction. Assume c=4.18 J/(g·K), ambient/initial 20 °C, reaction time 20 s, perfect mixing, constant heat generation rate. The loss is Newtonian cooling with dT/dt=q̇/C−0.020(T−20). The temperature at which the cooling curve is returned to the 10 s in the middle of the reaction is expressed as the highest extrapolation correction temperature. This approximation is not the actual maximum temperature and excludes actual reaction rates, evaporation, and thermometer delays.