0.897
J/(g·°C)
0.2144
cal/(g·°C)
5,000
J
5
kJ
—
°C
0.897
J/(g·°C)
0.2144
cal/(g·°C)
5,000
J
5
kJ
—
°C
The Specific Heat Calculator determines the specific heat capacity of a substance or solves related calorimetry problems using the equation c = q/(mΔT). Specific heat capacity is an intensive property that characterizes how much energy a material can absorb per unit mass per degree of temperature change. This property is unique to each substance, making it useful for material identification, thermal engineering, cooking science, and environmental modeling. The calculator accepts initial and final temperatures directly, converting to ΔT automatically, and provides results in both SI (J/(g·°C)) and calorie (cal/(g·°C)) units.
The specific heat capacity is defined as:
$$c = \frac{q}{m \cdot \Delta T}$$
where q = heat energy (J), m = mass (g), and ΔT = Tfinal − Tinitial (°C). Rearranging for other variables:
$$q = mc\Delta T, \quad m = \frac{q}{c\Delta T}, \quad \Delta T = \frac{q}{mc}$$
Unit conversions:
$$1 \text{ cal/(g·°C)} = 4.184 \text{ J/(g·°C)}$$
Water's specific heat (4.184 J/(g·°C) = 1.000 cal/(g·°C)) serves as the reference. Most metals have specific heats between 0.1 and 1 J/(g·°C), reflecting the Dulong-Petit law for molar heat capacities.
A higher specific heat means the substance requires more energy to change temperature — it is a better thermal buffer. Water (4.184) has one of the highest specific heats of any common substance, explaining its role in climate regulation. Metals like gold (0.129) and lead (0.128) have very low specific heats and heat up/cool down rapidly. If your calculated value matches a known substance, it helps identify the material.
Inputs
Results
c = 5000/(100 × 55) = 0.909 J/(g·°C) = 0.217 cal/(g·°C). Comparing with known values: aluminum = 0.897. Close match suggests the metal is likely aluminum.
Inputs
Results
q = 2000 × 0.449 × (200 − 25) = 2000 × 0.449 × 175 = 157,150 J = 157.15 kJ to heat a 2 kg iron skillet from 25°C to 200°C.
Specific heat capacity (c) is the amount of heat energy needed to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin). It is an intensive property — independent of sample size.
Water molecules form extensive hydrogen bonds. Heating water requires energy not only for molecular motion but also for partially breaking these hydrogen bonds, resulting in a high specific heat of 4.184 J/(g·°C).
The Dulong-Petit law states that the molar heat capacity of solid elements is approximately 25 J/(mol·K) or 3R. This means heavier elements have lower specific heats (per gram) since c ≈ 25/M.
Yes. Specific heat varies with temperature, especially near phase transitions. For precise calculations at extreme temperatures, use temperature-dependent Cp data. Near room temperature, values are approximately constant.
Cp is specific heat at constant pressure and Cv at constant volume. For solids and liquids, Cp ≈ Cv. For gases, Cp = Cv + R (per mole), where R = 8.314 J/(mol·K).
Specific heat is measured using calorimetry: a known mass of the substance is heated to a known temperature, placed in water, and the equilibrium temperature is measured. Using qmetal = −qwater solves for c.
Among common substances, hydrogen gas has the highest specific heat (14.3 J/(g·°C)). Among liquids, water (4.184) is highest. Among solids, lithium (3.58) is among the highest.
Metals have high atomic masses and metallic bonding allows relatively free energy distribution. By the Dulong-Petit law, molar heat capacity ≈ 25 J/(mol·K), so per-gram values decrease with increasing atomic mass.
No. During phase changes (melting, boiling), temperature is constant while heat is absorbed. Use q = mL (where L is latent heat) for phase transitions, and q = mcΔT only when temperature changes.
Standard specific heat values are known to 3-4 significant figures for common substances. Experimental determination in student labs typically achieves ±5-10% accuracy due to heat losses.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
How helpful was this calculator?
Be the first to rate!
