The heater warms more than the sample
Imagine a small metal sample sitting in water. An immersed resistor supplies electrical energy, and stirring keeps the water at nearly the same temperature throughout. The sample warms, but so do the water and the inner cup. A probe placed away from the resistor follows their temperature.
Dividing all the electrical energy by the sample’s temperature rise would assign the water’s and cup’s share to the sample. Even a light cup takes energy to warm. In a common double-cup calorimeter, the outer cup and air gap provide insulation. The inner aluminum cup belongs in the heat calculation.
Heat capacity tells us how much energy a particular object needs for one degree of warming. Specific heat tells us how much energy one kilogram needs for one degree of warming. A sample twice as heavy needs about twice as much energy for the same rise, though its specific heat remains approximately the same.
The electrical input comes from voltage, current and time. If the power changes, add the energy delivered in each short interval. Measurements of current and voltage throughout the run show how much energy the heater actually supplied.
Warm the same water and cup on their own
Remove the sample and repeat the experiment with the same cup, heater and mass of water. This is the blank. The heater stays immersed. The blank measures how much energy the apparatus already needs before the sample contributes anything.
Consider an ideal insulated example. The blank needs 280 joules for every degree of warming. With a 50-gram sample, the requirement becomes 325 joules per degree. The extra 45 joules per degree belongs to that sample. Dividing by its mass gives 900 joules per kilogram per degree, close to a common room-temperature value for aluminum.
With 1600 joules delivered in each run, the blank warms by about 5.71 degrees and the sample run by about 4.92 degrees. More heat capacity means less warming from the same energy. Keeping the water mass and vessel unchanged lets their contributions cancel when the two capacities are subtracted.
Keep watching after the power switches off
A real cup also loses energy to the room. As the bath gets warmer, heat generally escapes faster and the warming curve flattens. Assigning this escaped energy to the cup’s contents makes their inferred heat capacity too large.
Suppose the source supplies 1600 joules, but 200 joules leave the warm cup during heating. Only 1400 joules remain to raise the contents’ temperature. Using all 1600 joules as stored energy would make the inferred heat capacity about fourteen percent too large. The measured temperature rise belongs to the energy that stayed in the cup.
The cooling tail helps reveal that escape route. After the heater switches off, water warmer than the room cools toward room temperature. Faster heat transfer produces a faster fall. Using the warming and cooling portions together helps separate a large heat capacity from rapid heat loss.
The probe needs energy to warm too. When placed in warm water, it takes time to catch up. It can even keep warming briefly after the heater stops because it is still catching up with the bath. A separate record in a fixed-temperature bath measures this response before it is used to interpret the calorimeter’s curve.
These corrections suit a well-stirred bath over a modest temperature rise. If the sample’s center remains cold or substantial evaporation begins, one temperature cannot describe everything in the cup. The experiment then needs enough time and thermal contact for the material to warm evenly.