Science & engineering · Heat
Calorimetry calculator.
Mix a hot object into water, or solve for a metal's specific heat, assuming no heat loss.
Inputs
Result
Leave exactly one box empty.
How do you calculate the final temperature when two objects mix?
Heat is q = c x m x ΔT. In an ideal calorimeter the heat lost by the hotter body equals the heat gained by the cooler one, so m1 c1 (T1 − Tf) = m2 c2 (Tf − T2). For two bodies this gives Tf = (m1 c1 T1 + m2 c2 T2) / (m1 c1 + m2 c2). Water's specific heat is 4.184 J/g°C.
Worked example. A 360.0 g piece of steel (c about 0.449 J/g°C) dropped into 425 g of water at 24.0°C leaves them at 42.7°C. Solving for the start temperature gives about 248°C (OpenStax Chemistry 2e, 5.2).
Real calorimeters lose some heat and the cup absorbs some, so measured results differ from this ideal. Use your own measured specific heats if you have them.
Frequently asked questions
What is the calorimetry equation?
q = c m ΔT, where q is heat in joules, c is specific heat, m is mass and ΔT is the temperature change. With no heat loss, heat lost by one body equals heat gained by the other.
What is the specific heat of water?
4.184 J/g°C, the value used in OpenStax Chemistry 2e. Other textbooks round it to 4.18.
How do I find the specific heat of an unknown metal?
Heat the metal to a known temperature, drop it in water of known mass and temperature, measure the final temperature, and leave the body 1 specific heat box empty.
Why must the final temperature be between the two start temperatures?
Heat only flows from hot to cold in a mix, so the result cannot lie outside the two starting temperatures. The tool flags inputs that do.
Sources
Checked
Equation, water specific heat and the worked example are from the source above. Specific heat of steel (0.449) is the value used in that example. Runs entirely in your browser. Nothing you type is sent or stored.
Scope
Education reference. Assumes no heat lost to the container or surroundings and no phase change. Real experiments lose heat, so expect measured values to differ.