acalculator

What is the specific heat?

Pick what to find and type the other values. The specific heat calculator solves Q = m × c × ΔT for the specific heat capacity, the heat, the mass or the final temperature, with a table of common materials.

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Answer
c = 4186 J/(kg·°C)

The answer is c = 4186 J/(kg·°C).

Specific heat (J/(kg·°C))
4,186
Specific heat (J/(g·°C))
4.186
Specific heat (cal/(g·°C))
1.00048
Specific heat (Btu/(lb·°F))
0.999809
Heat (J)
41,860
Heat (kJ)
41.86
Heat (Btu)
39.6756
Mass (kg)
1
Temperature change (°C or K)
10
Temperature change (°F)
18
Final temperature (°C)
30
Final temperature (°F)
86

Answer: c = 4186 J/(kg·°C). The answer is c = 4186 J/(kg·°C).

How to calculate

Specific heat calculator: finds the specific heat capacity from heat, mass and temperature change (c = Q ÷ mΔT), or the heat, the mass or the final temperature from Q = mcΔT.

Example with the default inputs (Find Specific heat, Heat (Q) 39.6756246563155 BTU, Mass (m) 2.20462262184878 lb, Start temperature 67.9999999999999 °F, Final temperature 85.9999999999999 °F): The answer is c = 4186 J/(kg·°C).

Method: Q = m × c × ΔT, with ΔT = final − start temperature; c = Q ÷ (m × ΔT); m = Q ÷ (c × ΔT); final = start + Q ÷ (m × c).

  • The material stays in one phase (no melting, freezing or boiling), and c does not change with temperature over the range.
  • Table values are from OpenStax University Physics Volume 2, Table 1.3, in J/(kg·°C); a change of 1 °C is a change of 1 K.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Find Specific heat, Heat (Q) 39.68 BTU, Mass (m) 2.205 lb, Start temperature 68 °F, Final temperature 86 °F gives Specific heat (c) 4,186, Specific heat (cal/(g·°C)) 1.000478, Answer c = 4186 J/(kg·°C).Source: OpenStax, University Physics Volume 2, §1.4 Heat Transfer, Specific Heat, and Calorimetry (Q = mcΔT; Table 1.3), https://openstax.org/books/university-physics-volume-2/pages/1-4-heat-transfer-specific-heat-and-calorimetry (retrieved 2026-10-05)
  2. Find Heat, Material Aluminum, 900, Mass (m) 1.102 lb, Start temperature 68 °F, Final temperature 302 °F gives Heat (J) 58,500, Heat (kJ) 58.5.Source: OpenStax, University Physics Volume 2, §1.4 Heat Transfer, Specific Heat, and Calorimetry (Q = mcΔT; Table 1.3), https://openstax.org/books/university-physics-volume-2/pages/1-4-heat-transfer-specific-heat-and-calorimetry (retrieved 2026-10-05) (c of aluminum, Table 1.3)
  3. Find Final temperature, Material Water (liquid), 4,186, Heat (Q) 19.84 BTU, Mass (m) 0.5512 lb, Start temperature 59 °F gives Temperature change (°C or K) 20, Final temperature (°C) 35.Source: OpenStax, University Physics Volume 2, §1.4 Heat Transfer, Specific Heat, and Calorimetry (Q = mcΔT; Table 1.3), https://openstax.org/books/university-physics-volume-2/pages/1-4-heat-transfer-specific-heat-and-calorimetry (retrieved 2026-10-05)
  4. Find Mass, Material Other: type the specific heat, Specific heat (c) 1, Specific heat unit Btu/(lb·°F), Heat (Q) 1 BTU, Start temperature 68 °F, Final temperature 69 °F gives Mass (kg) 0.453592, Specific heat (c) 4,186.8.Source: NIST SP 811, Appendix B.8 (1 Btu_IT = 1,055.05585262 J; 1 cal_th = 4.184 J; 1 lb = 0.45359237 kg), https://www.nist.gov/pml/special-publication-811 (retrieved 2026-10-05)

How it works

The heat Q that changes the temperature of a mass m of one material is

  • Q = m × c × ΔT, where ΔT = final temperature − start temperature.

Rearranged for each unknown:

  • Specific heat c = Q ÷ (m × ΔT)
  • Heat Q = m × c × ΔT
  • Mass m = Q ÷ (c × ΔT)
  • Final temperature = start temperature + Q ÷ (m × c)

Materials (OpenStax Table 1.3, J/(kg·°C)): water 4,186; ice 2,090; aluminum 900; copper 387; iron or steel 452; lead 128; silver 235; gold 129; glass 840; concrete or granite 840; wood 1,700; ethanol 2,450; mercury 139. Or type your own c in J/(kg·°C), kJ/(kg·°C), J/(g·°C), cal/(g·°C) or Btu/(lb·°F).

Unit sizes (exact): 1 kJ/(kg·°C) = 1 J/(g·°C) = 1,000 J/(kg·°C); 1 cal/(g·°C) = 4,184 J/(kg·°C); 1 Btu/(lb·°F) = 1,055.05585262 ÷ (0.45359237 × 5/9) = 4,186.8 J/(kg·°C). Temperatures: K = °C + 273.15 = (°F + 459.67) × 5/9.

Rules

  • Heat is from −10¹⁵ to 10¹⁵ J (negative means heat taken away); mass more than 0 and at most 10⁹ kg; a typed c more than 0 and at most 10⁶ in its unit; temperatures from 0 K to 100,000 K.
  • Finding c or m: the start and final temperatures must differ, and Q must have the same sign as ΔT (heat in warms, heat out cools). Otherwise there is no answer.
  • Finding Q: no answer when |Q| is over 10¹⁵ J. Finding m: no answer when m is over 10⁹ kg.
  • Finding the final temperature: no answer when it falls below 0 K or rises above 100,000 K.

Output format

The answer line shows the found value to 6 significant figures: c in J/(kg·°C), Q in J and Btu, m in kg and lb, or the final temperature in °C and °F. The list shows c in four units, Q in J, kJ and Btu, m in kg, ΔT in °C and °F, and the final temperature in °C and °F, each to 6 significant figures. Typed values are read back as exact decimals and the arithmetic is exact; each value is rounded once. The answer line writes each number as a plain decimal from 10⁻⁶ up to 10²¹ (else d.ddde+N), rounded half up from its exact value, with the true minus sign (−) for a negative value.

Worked examples by hand

Water. 41,860 J warms 1 kg from 20 °C to 30 °C: c = 41,860 ÷ (1 × 10) = 4,186 J/(kg·°C) = 4,186 ÷ 4,184 = 1.00048 cal/(g·°C).

An aluminum pan. 0.500 kg of aluminum (900 J/(kg·°C)) from 20 °C to 150 °C: Q = 0.5 × 900 × 130 = 58,500 J = 58.5 kJ.

Final temperature. 20,930 J into 0.25 kg of water at 15 °C: ΔT = 20,930 ÷ (0.25 × 4,186) = 20 °C, so the water ends at 35 °C.

Mass in US units. 1 Btu warms a material of 1 Btu/(lb·°F) by 1 °F: c = 4,186.8 J/(kg·°C), ΔT = 5/9 °C, so m = 1,055.05585262 ÷ (4,186.8 × 5/9) = 0.45359237 kg, which is 1 lb.

Other questions people ask

How do you calculate specific heat?

Divide the heat by the mass times the temperature change: c = Q ÷ (m × ΔT). If 41,860 J warms 1 kg by 10 °C, c = 41,860 ÷ (1 × 10) = 4,186 J/(kg·°C), the value for water.

What is the specific heat of water?

About 4,186 J/(kg·°C), or 4.186 J/(g·°C), or 1.000 cal/(g·°C), or 1.000 Btu/(lb·°F). It takes 4,186 J to warm 1 kg of liquid water by 1 °C.

What does Q = mcΔT mean?

Q is the heat that goes in (positive) or out (negative), m is the mass, c is the specific heat of the material, and ΔT is the final temperature minus the start temperature.

Is ΔT the same in Celsius and kelvins?

Yes. A change of 1 °C is a change of 1 K, so J/(kg·°C) and J/(kg·K) are the same unit. A change of 1 °F is 5/9 of that.

Why does a negative heat give a lower final temperature?

A negative Q is heat taken away, so the material cools. When you find c or m, Q and ΔT must have the same sign, because c and m are always positive.

How much heat does it take to warm an aluminum pan?

A 0.500 kg aluminum pan from 20 °C to 150 °C needs 0.500 × 900 × 130 = 58,500 J, or 58.5 kJ, with c = 900 J/(kg·°C).

Does this work when water boils or ice melts?

No. Q = mcΔT holds only while the material stays in one phase. Melting and boiling take latent heat at a fixed temperature, which this formula does not count.