What is the specific heat formula?

The standard formula is Q = m × c × ΔT, where Q is the heat energy transferred (J), m is the mass of the substance (g), c is the specific heat capacity (J g⁻¹ K⁻¹) and ΔT is the temperature change in kelvin or °C (they have the same magnitude). Rearranging gives m = Q ÷ (c × ΔT), c = Q ÷ (m × ΔT) and ΔT = Q ÷ (m × c).

What is the specific heat capacity of water?

Liquid water has a specific heat capacity of 4.184 J g⁻¹ K⁻¹ (also written 4,184 J kg⁻¹ K⁻¹ or 1 cal g⁻¹ °C⁻¹) at 25 °C and 1 atm. This exceptionally high value — roughly 10 times that of copper — is why water is used as a coolant and why oceans moderate coastal climates.

What is the difference between specific heat capacity and heat capacity?

Heat capacity (symbol C, unit J K⁻¹) is a property of a specific sample — it tells you how many joules that exact sample needs per kelvin of warming. Specific heat capacity (symbol c, unit J g⁻¹ K⁻¹ or J kg⁻¹ K⁻¹) is normalised per unit mass, making it a material property that applies to any size sample. Multiplying specific heat by mass gives heat capacity: C = m × c.

Why is ΔT the same in °C and K?

The Celsius and Kelvin scales differ only by an offset of 273.15 — every degree Celsius is exactly 1 kelvin wide. So a change of 10 °C is the same as a change of 10 K. When you use ΔT in the formula Q = m·c·ΔT, only the difference matters, not the absolute value, so °C and K are interchangeable for ΔT.

How do I convert °F temperature changes to kelvin?

A Fahrenheit degree is 5/9 the size of a kelvin. To convert a temperature difference from °F to K, multiply by 5/9. For example, a change of 18 °F = 18 × 5/9 = 10 K. The calculator does this conversion automatically when you select the °F unit for ΔT.

What is calorimetry and how does this calculator help?

Calorimetry is the science of measuring heat exchange during chemical reactions, phase changes and physical processes. A simple coffee-cup calorimeter measures the temperature change of a known mass of water; you can then use Q = m·c·ΔT to find the heat released or absorbed. This calculator automates that calculation and lets you rearrange for any unknown, saving time in lab reports and homework.

Specific Heat Calculator

A specific heat calculator that applies the fundamental thermodynamic equation Q = m × c × ΔT to find any one of the four variables — heat energy Q, mass m, specific heat capacity c, or temperature change ΔT — given the other three. It covers 15 common materials (water, ice, steam, ethanol, metals, gases) with NIST-verified capacities, supports six energy units (J, kJ, cal, kcal, Wh, BTU), five mass units (g, kg, mg, lb, oz) and both K/°C and °F temperature differences.

The formula

The equation connecting heat transfer to mass and temperature is:

Q = m · c · ΔT

Symbol	Quantity	SI unit
Q	Heat energy transferred	Joule (J)
m	Mass of the substance	gram (g) or kilogram (kg)
c	Specific heat capacity	J g⁻¹ K⁻¹ or J kg⁻¹ K⁻¹
ΔT	Change in temperature = T_final − T_initial	kelvin (K) or °C

Because ΔT in °C equals ΔT in K (the two scales have identical step sizes), you can use either without conversion as long as you are consistent. If you prefer Fahrenheit differences, multiply by 5/9 to get the kelvin equivalent.

Rearranging for each unknown:

m = Q ÷ (c × ΔT) — find the mass needed to absorb a given amount of heat
c = Q ÷ (m × ΔT) — identify a material from a calorimetry measurement
ΔT = Q ÷ (m × c) — predict how much a substance will warm or cool

How it works

The calculator converts all inputs to a coherent base (joules, grams, kelvin), applies the appropriate rearrangement, then converts the result back to your chosen output unit. For Fahrenheit temperature differences the conversion factor 5/9 is applied before the main calculation (and 9/5 on the result when solving for ΔT). All 15 preset specific heat values come from the CRC Handbook of Chemistry and Physics (103rd edition) and the NIST Chemistry WebBook. Nothing is sent to a server — every calculation runs locally in your browser.

Worked example — heating water for coffee

Suppose you want to heat 250 g of water from 20 °C to 100 °C. How much heat energy is required?

Given: m = 250 g, c = 4.184 J g⁻¹ K⁻¹ (liquid water), ΔT = 100 − 20 = 80 K

Step 1: Apply the formula: Q = 250 g × 4.184 J g⁻¹ K⁻¹ × 80 K

Step 2: Calculate: Q = 83,680 J = 83.68 kJ ≈ 20.0 kcal

So you need about 83.7 kJ to boil water for a single cup of coffee. A 3 kW kettle would deliver this in roughly 28 seconds of heating time.

Worked example — identifying a metal (calorimetry)

A 50 g sample of an unknown metal absorbs 1,540 J and its temperature rises by 68 °C. What is its specific heat capacity, and what metal might it be?

Given: Q = 1,540 J, m = 50 g, ΔT = 68 K

Step: c = Q ÷ (m × ΔT) = 1,540 ÷ (50 × 68) = 0.453 J g⁻¹ K⁻¹

Comparing with the reference table — iron/steel is 0.449 J g⁻¹ K⁻¹ — the sample is almost certainly iron or steel. This is the standard calorimetry workflow used in undergraduate chemistry labs.

Why specific heat matters

Water’s high specific heat capacity (4.184 J g⁻¹ K⁻¹) explains why oceans and lakes moderate coastal climates: they absorb enormous amounts of solar energy while rising only a few degrees. Metals have much lower values — lead absorbs about 33 times less heat per gram per kelvin than water — which is why metal objects feel “colder” to the touch even at the same room temperature (they conduct heat away faster and heat up quickly). Engineers use the formula when sizing cooling systems, HVAC units, heat exchangers and thermal batteries.