Entropy Change Calculator
Compute the entropy change of a substance undergoing a constant-pressure heating or cooling process. Used in thermodynamics courses and engineering analyses of heat exchangers, turbines, and compressors.
About this calculator
For a substance with constant specific heat capacity undergoing a reversible constant-pressure process, the entropy change is given by ΔS = m × c_p × ln(T_f / T_i), where m is the mass in kilograms, c_p is the specific heat capacity in J/kg·K, T_f is the final absolute temperature in Kelvin, and T_i is the initial absolute temperature in Kelvin. The natural logarithm (ln) appears because entropy is defined through the integral of dQ/T over the process path. A positive ΔS indicates entropy has increased (the substance absorbed heat), while a negative value means entropy decreased (heat was removed). This formula assumes c_p is constant over the temperature range, a valid approximation for many solids and liquids. For gases with strongly temperature-dependent heat capacities, tabulated entropy data is more accurate.
How to use
Consider 5 kg of water (c_p = 4,186 J/kg·K) heated from 300 K to 400 K at constant pressure. Enter 5 in Mass, 4186 in Specific Heat Capacity, 300 in Initial Temperature, and 400 in Final Temperature. The calculator computes ΔS = 5 × 4186 × ln(400 / 300) = 20,930 × ln(1.3333) = 20,930 × 0.2877 ≈ 6,021 J/K. The positive result confirms entropy increased as heat was added to the water. This result can be used to verify the thermodynamic feasibility of the heating process.
Frequently asked questions
Why does the entropy change formula use a natural logarithm instead of a simple temperature difference?
Entropy is defined as the integral of reversible heat transfer divided by absolute temperature: dS = dQ_rev / T. For a constant-pressure process with constant c_p, integrating dQ = m × c_p × dT from T_i to T_f yields ΔS = m × c_p × ∫(dT/T) = m × c_p × ln(T_f / T_i). The natural logarithm arises directly from this integration of 1/T. Using a simple temperature difference (T_f − T_i) would be mathematically incorrect and would not properly account for the fact that heat added at higher temperatures contributes less entropy change than the same heat added at lower temperatures.
What units should I use for temperature when calculating entropy change?
Temperatures must be entered in Kelvin (absolute temperature scale) for the entropy formula to work correctly. The ratio T_f / T_i is dimensionless, so both temperatures must be on the same absolute scale. Using Celsius values would produce a wrong ratio — for instance, heating from 27 °C to 127 °C should use 300 K and 400 K, giving ln(400/300) ≈ 0.288, whereas using Celsius would give ln(127/27) ≈ 1.55, a completely different and incorrect result. Always add 273.15 to Celsius values before entering them.
Can this entropy change calculator be used for gases as well as liquids and solids?
Yes, the formula ΔS = m × c_p × ln(T_f / T_i) applies to gases at constant pressure, provided the specific heat capacity c_p is treated as constant. For ideal gases, c_p is approximately constant over moderate temperature ranges — for example, air has c_p ≈ 1,005 J/kg·K near ambient conditions. However, for wide temperature swings or gases like CO₂ with strongly temperature-dependent heat capacities, the constant c_p assumption introduces error. In those cases, engineers use tabulated entropy values (e.g., from the JANAF thermochemical tables) or integrate a temperature-dependent c_p expression for greater accuracy.