15,690
15,690
J
15.69
kJ
0
g
0
J/(g·°C)
15
°C
15,690
J
15,690
15,690
J
15.69
kJ
0
g
0
J/(g·°C)
15
°C
15,690
J
The Calorimetry Calculator solves the fundamental calorimetry equation q = mcΔT for any of its four variables: heat (q), mass (m), specific heat capacity (c), or temperature change (ΔT). Calorimetry is the science of measuring heat transfer during chemical reactions or physical changes. This versatile calculator handles scenarios from coffee-cup calorimetry experiments to industrial heat exchange calculations. Whether you need to find how much heat is required to warm water, identify a metal by its specific heat, or determine the temperature change from a known energy input, this calculator provides instant results with proper unit handling.
The fundamental calorimetry equation relates four quantities:
$$q = mc\Delta T$$
where:
Solving for each variable:
$$m = \frac{q}{c \cdot \Delta T}, \quad c = \frac{q}{m \cdot \Delta T}, \quad \Delta T = \frac{q}{m \cdot c}$$
Sign convention: positive q means heat absorbed (endothermic), negative q means heat released (exothermic). ΔT = Tfinal − Tinitial.
When solving for heat: a positive result means the substance absorbed heat (warmed up), negative means it released heat. For mass and specific heat: results should always be positive. A calculated specific heat can help identify unknown substances by comparing with known values. For temperature change: the sign tells you direction — positive means warming, negative means cooling.
Inputs
Results
q = 250 × 4.184 × 15 = 15,690 J = 15.69 kJ. This is the energy needed to heat a cup of water from room temperature to about 37°C.
Inputs
Results
c = 548/(50 × 28) = 0.391 J/(g·°C). Comparing with known values: Cu = 0.385, this is likely copper.
Calorimetry is the measurement of heat transfer during chemical reactions or physical changes. It uses the relationship q = mcΔT to quantify energy changes in a system.
Specific heat capacity (c) is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1°C. Water has c = 4.184 J/(g·°C), which is unusually high.
Specific heat (c) is per gram: units of J/(g·°C). Heat capacity (C) is for the entire sample: C = mc, with units of J/°C. Heat capacity depends on sample size; specific heat does not.
Water: 4.184, ice: 2.09, steam: 2.01, aluminum: 0.897, iron: 0.449, copper: 0.385, gold: 0.129 J/(g·°C). Metals generally have low specific heats.
Yes. Since a change of 1°C equals a change of 1 K, ΔT has the same numerical value in both scales. However, do not confuse absolute temperature with temperature change.
A simple constant-pressure calorimeter made from insulated cups. It assumes all heat exchange occurs between the reaction and the solution (usually water), with negligible heat loss to surroundings.
Real calorimeters lose some heat to surroundings. You can minimize this with insulation, or correct for it using the calorimeter constant (determined by calibration) and extrapolation techniques.
No. During phase changes (melting, boiling), temperature remains constant while heat is absorbed. Use q = nΔHfus or q = nΔHvap instead. The q = mcΔT formula only applies when temperature changes.
In an isolated system, heat lost by the hot object equals heat gained by the cold object: qhot + qcold = 0, or m₁c₁ΔT₁ = −m₂c₂ΔT₂. This is used to find equilibrium temperatures.
Water's high specific heat is due to extensive hydrogen bonding between molecules. Breaking and forming these hydrogen bonds absorbs significant energy, buffering temperature changes.
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