Theory — Entropy Change for Heating or Cooling at Constant Pressure
For a reversible temperature change of a closed system at constant pressure,
Clausius’ definition \( \mathrm dS = \delta q_{\mathrm{rev}}/T \) together with
\( \delta q_{\mathrm{rev}} = C_p\,\mathrm dT \) leads to
\[
\begin{aligned}
\Delta S
&= \int_{T_i}^{T_f}\frac{C_p}{T}\,\mathrm dT \\
&\approx C_p \,\ln\!\left(\frac{T_f}{T_i}\right)
\end{aligned}
\]
The logarithmic form holds when \(C_p\) is effectively constant over the interval
\([T_i,\,T_f]\). If \(C_p\) varies appreciably with \(T\), evaluate the exact integral
\(\Delta S=\int (C_p(T)/T)\,\mathrm dT\) or split the path into smaller temperature segments.
Basis and units
- Molar basis: \(C_{p,m}\) in \(\mathrm{J\,mol^{-1}\,K^{-1}}\).
For \(n\) moles, the total change is
\(\Delta S_{\text{total}} = n\,C_{p,m}\ln(T_f/T_i)\) (\(\mathrm{J\,K^{-1}}\)).
- Mass basis: \(c_p\) in \(\mathrm{J\,kg^{-1}\,K^{-1}}\).
For mass \(m\), \(\Delta S_{\text{total}} = m\,c_p\ln(T_f/T_i)\).
- Temperatures must be on the absolute (kelvin) scale.
Rearrangements used by the calculator
\[
\begin{aligned}
\Delta S_{\text{total}}
&= \big(C_p^\ast \,\text{amount}\big)\,\ln\!\left(\frac{T_f}{T_i}\right) \\
C_p^\ast
&= \frac{\Delta S_{\text{total}}}{(\text{amount})\,\ln(T_f/T_i)} \\
T_f
&= T_i\,\exp\!\left(\frac{\Delta S_{\text{total}}}{C_p^\ast\,\text{amount}}\right),
\qquad
T_i
&= T_f\,\exp\!\left(-\frac{\Delta S_{\text{total}}}{C_p^\ast\,\text{amount}}\right)
\end{aligned}
\]
Here \(C_p^\ast\) is the heat capacity per chosen basis (per mole or per kg);
“amount” is \(n\) (mol) or \(m\) (kg).
Signs and qualitative trends
- If \(T_f>T_i\): \(\ln(T_f/T_i)>0\) so \(\Delta S>0\) (heating increases entropy).
- If \(T_f<T_i\): \(\ln(T_f/T_i)<0\) so \(\Delta S<0\) (cooling decreases entropy).
Worked example
Heat \(1.00\ \mathrm{mol}\) of liquid water from \(298\ \mathrm{K}\) to \(373\ \mathrm{K}\)
using \(C_{p,m} \approx 75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}\) (assumed constant).
\[
\begin{aligned}
\Delta S_{\text{total}}
&= (1.00\ \mathrm{mol})\,(75.3\ \mathrm{J\,mol^{-1}\,K^{-1}})
\,\ln\!\left(\frac{373}{298}\right) \\
&= 75.3 \times 0.224 \\
&\approx 16.9\ \mathrm{J\,K^{-1}},
\qquad
\Delta S_{\mathrm m} \approx 16.9\ \mathrm{J\,mol^{-1}\,K^{-1}}
\end{aligned}
\]
When the formula should not be used directly
- The path crosses a phase transition. Treat the path in segments and include the isothermal step
with \( \Delta_{\mathrm{tr}}S = \Delta_{\mathrm{tr}}H/T_{\mathrm{tr}} \) at the transition.
- \(C_p\) varies strongly over the interval (very wide ranges or near critical points).
Use \(C_p(T)\) and integrate, or apply piecewise over smaller intervals.
Common pitfalls
- Convert any Celsius temperatures to kelvin before taking the ratio:
\(T(\mathrm K) = T(^{\circ}\mathrm C)+273.15\).
- Use the natural logarithm \(\ln\), not \(\log_{10}\).
- Report units consistent with your basis (per mole vs per kg) when quoting \(\Delta S\).
