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Accurate heat‑transfer experiments hinge on a well‑calibrated calorimeter. Whether you’re using a simple Styrofoam cup or a sophisticated, explosion‑proof vessel, the container absorbs a portion of the heat released or absorbed during a reaction. By determining the calorimeter constant— the amount of energy needed to raise the calorimeter’s temperature by 1 °C—you can correct for this loss and measure the specific heat of unknown substances with confidence.
The most reliable calibration method involves mixing two equal masses of water at different temperatures inside the calorimeter and recording the equilibrium temperature. Water’s specific heat (Cs) is 1 cal g⁻¹ °C⁻¹ (4.186 J g⁻¹ °C⁻¹), making it an ideal calibration fluid.
Let a known mass of hot water (m1) be poured into a calorimeter that already contains a known mass of cold water (m2). Measure the initial temperatures T1 and T2, then wait for the mixture to reach the equilibrium temperature TE. The temperature changes are ΔT1 = TE – T1 and ΔT2 = TE – T2 (note that ΔT2 is positive because the cold water warms).
The heat lost by the hot water is q1 = m1CsΔT1, while the heat gained by the cold water is q2 = m2CsΔT2. The difference, q1 – q2, is the heat absorbed by the calorimeter:
qabs = (m1CsΔT1) – (m2CsΔT2)
Because the calorimeter’s temperature rise equals ΔT2, its heat capacity (the calorimeter constant, cc) is:
cc = qabs ÷ ΔT2 = Cs(m1ΔT1 + m2ΔT2) ÷ ΔT2 cal g⁻¹ °C⁻¹
With cc known, you can determine the specific heat of any material. Heat a known mass of the substance (m1) to temperature T1 and place it in the calorimeter containing an equal mass (m2) at a cooler temperature T2. After the system reaches equilibrium at TE, calculate ΔT1 = TE – T1 and ΔT2 = TE – T2.
Rearranging the energy balance gives the substance’s specific heat:
Cs = (cc × ΔT2) ÷ (m1ΔT1 + m2ΔT2) cal g⁻¹ °C⁻¹
Converting to joules is straightforward: multiply the result by 4.186 J cal⁻¹.
