Heat & Heat Capacity — \(q=C\,\Delta T\)
When a sample warms or cools without a phase change or reaction, the heat exchanged is
\( q = C\,\Delta T \). The total (bulk) heat capacity is \(C=m\,c_p\) if using a
specific heat capacity \(c_p\) (per gram), or \(C=n\,C_{p,m}\) if using a
molar heat capacity \(C_{p,m}\) (per mole).
\[
\begin{aligned}
q &= C\,\Delta T \\
C &= m\,c_p \quad \text{or} \quad C = n\,C_{p,m}
\end{aligned}
\]
Key relations & unit links
Specific ↔ molar heat capacity convert via the molar mass \(M\) (g·mol\(^{-1}\)):
\[
\begin{aligned}
C_{p,m} &= c_p \cdot M \\
c_p &= \dfrac{C_{p,m}}{M}
\end{aligned}
\]
Typical units: \(c_p\) in \(\mathrm{J\,g^{-1}}\,^{\circ}\mathrm{C}^{-1}\),
\(C_{p,m}\) in \(\mathrm{J\,mol^{-1}}\,^{\circ}\mathrm{C}^{-1}\), and \(C\) in
\(\mathrm{J}\,^{\circ}\mathrm{C}^{-1}\). Use consistent units (e.g., mass in g if \(c_p\) is per g).
Why \(\Delta T\) is the same in °C and K
The temperature difference is identical in kelvin and in degrees Celsius:
\[
\begin{aligned}
T(\mathrm{K}) &= T(^{\circ}\mathrm{C}) + 273.15 \\
\Delta T_{\mathrm{K}} &= \big(T_{f,^{\circ}\mathrm{C}}+273.15\big) - \big(T_{i,^{\circ}\mathrm{C}}+273.15\big) \\
&= T_{f,^{\circ}\mathrm{C}} - T_{i,^{\circ}\mathrm{C}} \;=\; \Delta T_{^{\circ}\mathrm{C}}
\end{aligned}
\]
Sign convention & energy conservation
Heating the system gives \(q>0\); cooling gives \(q<0\). System and surroundings obey energy conservation:
\[
\begin{aligned}
\Delta T>0 &\;\Rightarrow\; q>0 \quad (\text{system absorbs heat})\\
\Delta T<0 &\;\Rightarrow\; q<0 \quad (\text{system releases heat})\\[4pt]
q_{\text{system}} + q_{\text{surroundings}} &= 0
\end{aligned}
\]
For a two-body mixing/heating scenario at constant pressure (no losses):
\[
m_1 c_{p,1}\,\big(T_f - T_{i,1}\big) \;+\; m_2 c_{p,2}\,\big(T_f - T_{i,2}\big) \;=\; 0.
\]
Worked example (matches the example button)
Heat needed to warm \(100\ \mathrm{g}\) of liquid water from \(21.0\,^{\circ}\mathrm{C}\) to \(37.0\,^{\circ}\mathrm{C}\)
using \(c_p(\text{water, l}) \approx 4.184\ \mathrm{J\,g^{-1}}\,^{\circ}\mathrm{C}^{-1}\):
\[
\begin{aligned}
\Delta T &= 37.0 - 21.0 = 16.0\,^{\circ}\mathrm{C} \\
C &= m\,c_p \;=\; 100\,\mathrm{g}\,\cdot\,4.184\ \mathrm{J\,g^{-1}}\,^{\circ}\mathrm{C}^{-1}
\;=\; 418.4\ \mathrm{J}\,^{\circ}\mathrm{C}^{-1} \\
q &= C\,\Delta T \;=\; 418.4\,\cdot\,16.0
\;=\; 6.69 \times 10^{3}\ \mathrm{J}
\end{aligned}
\]
Constant-pressure vs constant-volume (context)
\(c_p\) and \(c_v\) differ for gases; for an ideal gas (per mole):
\[
C_{p,m} - C_{v,m} \;=\; R.
\]
This calculator uses \(q=C\,\Delta T\) with the capacity you provide. At constant pressure, use \(c_p\) or \(C_{p,m}\);
at constant volume, use \(c_v\) or \(C_{v,m}\) if appropriate.
Validity & common pitfalls
- No phase change: If the path crosses a phase change, use latent heat \(q = m\,L\) for that segment and sum piecewise.
- Capacity nearly constant: \(c_p\) (or \(C_{p,m}\)) should not vary strongly over the temperature range used.
- Unit consistency: Match mass units to \(c_p\) units and temperature difference units to those of \(c_p\)/\(C_{p,m}\).
- Sign awareness: Cooling (\(\Delta T<0\)) yields \(q<0\); report magnitude and sign correctly.
