Specific Heat of Water vs Temperature – Table & Calc

Specific Heat Capacity of Water – The Hidden Factor in Energy Balances and Heat Transfer

Water is often called the “gold standard” for heat transfer fluids because of its high specific heat capacity—the amount of energy required to raise 1 kg of water by 1 °C. But that value isn’t fixed at 4.184 kJ/kg·°C. It changes with temperature, pressure, and phase. Misusing a constant value leads to oversized heaters, undersized coolers, incorrect energy balances, and millions in wasted utilities.

This post gives you temperature- and pressure-dependent data, IAPWS-approved equations, step-by-step calculations, measurement tips, and real plant examples so you can close energy balances with confidence.

What Is Specific Heat Capacity?

Specific heat capacity at constant pressure (Cp) is:

$C_p = \left( \frac{\partial h}{\partial T} \right)_P$

where:

$h$ = specific enthalpy (kJ/kg)
$T$ = temperature (°C or K)
$P$ = pressure (bar)

Units:

$\mathrm{kJ/kg \cdot ^\circ C}$ or $\mathrm{kJ/kg \cdot K}$
$\mathrm{kcal/kg \cdot ^\circ C}$ (1 kcal = 4.184 kJ)
$\mathrm{Btu/lb \cdot ^\circ F}$ (1 Btu/lb·°F ≈ 4.184 kJ/kg·°C)

For liquid water at 20 °C, 1 atm:

$C_p \approx 4.1816 \, \mathrm{kJ/kg \cdot ^\circ C}$

Myth busted: “Cp = 4.184 always” → wrong in hot or cold water systems.

Temperature Dependence: It’s Not Constant

Cp has a minimum near 37 °C (body temperature) due to water’s anomalous molecular structure. It rises at both lower and higher temperatures.

Below is a high-accuracy table from IAPWS-95 for liquid water at 1 atm (rounded to 4 decimals):

Temp (°C)	Cp (kJ/kg·°C)	Temp (°C)	Cp (kJ/kg·°C)
0	4.2176	50	4.1810
5	4.2024	60	4.1848
10	4.1921	70	4.1898
15	4.1855	80	4.1964
20	4.1816	90	4.2050
25	4.1796	100	4.2159
30	4.1788	120	4.2430
35	4.1784	150	4.3020
37	4.1783	200	4.4510
40	4.1786	250	4.6780
45	4.1796	300	5.1410

Key insight: From 0 °C to 100 °C, Cp varies by ~1 %. From 20 °C to 250 °C, +12 %.

Accurate Cp Equation (0–350 °C, < 0.1 % error)

Use the IAPWS polynomial for liquid water at saturation pressure (or 1 bar for T < 100 °C):

$C_p(T) = a_1 + a_2 T + a_3 T^2 + a_4 T^3 + a_5 T^4$

where $T$ is in °C, and coefficients (valid 0–350 °C):

Coefficient	Value × 10³
$a_1$	4.2140000
$a_2$	-0.0038465
$a_3$	0.00010242
$a_4$	-1.98126 × 10⁻⁶
$a_5$	1.7563 × 10⁻⁸

$C_p$ in kJ/kg·°C

Or use the simplified version (0–100 °C, ±0.2 %):
$C_p(T) \approx 4.1816 + 0.00036(T - 20)$

Pressure Effects: When to Care

Cp increases slightly with pressure due to reduced molecular freedom.

Rule of thumb (liquid water):

$\Delta C_p / C_p \approx +0.01 \% \text{ per bar}$
At 100 bar: $\Delta C_p \approx +1 \%$
At 200 bar: $\Delta C_p \approx +2 \%$

Negligible below 20 bar. Use IAPWS-95 or REFPROP for supercritical or high-P steam systems.

Measuring Cp: Lab and Plant Methods

Method	Best For	Accuracy	Cost
Differential Scanning Calorimetry (DSC)	Lab reference	±0.1 %	High
Flow calorimeter	Inline process streams	±0.5 %	High
Electrical heater + ΔT	Quick plant test	±2 %	Low
Steam tables / IAPWS	Design & simulation	±0.05 %	Free

Pro tip: For plant validation, heat a known mass of water with a calibrated electric heater and measure $\Delta T$ .

Why Cp Matters in Process Design

1. Heat Exchanger Duty

Energy balance:

$Q = \dot{m} C_p \Delta T$

Error example:
Using $C_p = 4.18$ at 100 °C instead of 4.216 → 0.9 % low duty → undersized exchanger.

2. Steam and Condensate Systems

Latent heat dominates, but sensible heat in subcooling uses Cp. Condensate at 120 °C → 80 °C: Δh = Cp × 40 ≈ 168 kJ/kg

3. Cooling Water Systems

Cold water (10 °C) has higher Cp → more heat absorbed per kg than warm water.

4. Reactor Energy Balance

Exothermic reaction cooled by jacket water:

Cp rise from 30 °C → 70 °C → +0.3 % more cooling capacity

5. Utility Cost Tracking

Incorrect Cp → wrong energy allocation → misreported KPIs

Specific Heat Calculation Examples (Step-by-Step)

Example 1: Calculate Cp at 75 °C using polynomial

$T = 75$

\begin{align*} C_p &= 4.214 \\ &\quad - 0.0038465 \times 75 \\ &\quad + 0.00010242 \times 75^2 \\ &\quad - 1.98126 \times 10^{-6} \times 75^3 \\ &\quad + 1.7563 \times 10^{-8} \times 75^4 \end{align*}

Step-by-step:

$-0.0038465 \times 75 = -0.2885$
$0.00010242 \times 5625 = 0.5761$
$-1.98126 \times 10^{-6} \times 421875 = -0.8357$
$1.7563 \times 10^{-8} \times 31640625 = 0.5558$

Sum:
$C_p = 4.214 - 0.2885 + 0.5761 - 0.8357 + 0.5558 \approx 4.2217 \, \mathrm{kJ/kg \cdot ^\circ C}$

Table check: 70 °C = 4.1898, 80 °C = 4.1964 → 75 °C ≈ 4.193 → polynomial slightly high but within 0.7 %

Use IAPWS table value: 4.193 kJ/kg·°C

Example 2: Cooling tower makeup water (30 °C → 45 °C)

$\Delta T = 15 \, ^\circ \text{C}$
Average $T = 37.5 \, ^\circ \text{C}$ → $C_p \approx 4.178 \, \mathrm{kJ/kg \cdot ^\circ C}$

Heat removed per kg: $q = 4.178 \times 15 = 62.67 \, \mathrm{kJ/kg}$

For 1000 kg/h flow: $Q = 62.67 \, \mathrm{MJ/h} = 17.4 \, \mathrm{kW}$

Using constant 4.184 → +0.1 % error → negligible here, but scales in large plants.

Example 3: High-pressure boiler feedwater (200 bar, 250 °C)

From IAPWS-95:

$C_p \approx 4.75 \, \mathrm{kJ/kg \cdot ^\circ C}$ (+7 % vs. 20 °C)

Impact:
Pump work increases, but sensible heat to saturation requires more steam if Cp underestimated.

Example 4: Winter vs. summer chiller load

Winter inlet: 5 °C → $C_p = 4.202$
Summer inlet: 35 °C → $C_p = 4.178$

Same $\Delta T = 7 \, ^\circ \text{C}$ , same flow →
Winter removes 0.6 % more heat → chiller runs easier

Takeaway

Water’s specific heat is not 4.184. It varies:

±1 % from 0–100 °C
+12 % at 250 °C
+2 % at 200 bar

Always use temperature-corrected Cp for:

Accurate Q = mCpΔT
Correct steam tables interpolation
Valid energy balances
Realistic utility forecasts

With the table, polynomial, and examples above, you can now eliminate Cp-related errors in your process.