Specific Heat Calculator

Results

How to Use the Specific Heat Calculator

1. 1. Choose what to calculate. Use the first option for q = mcΔT, the second for the final temperature of a hot object dropped into water, the third for melting, freezing, boiling or condensing, and the fourth to total the heat for ice becoming steam.
2. 2. Pick the unknown and enter the rest. Energy can be J, kJ, cal, kcal or Btu; mass from mg to lb; specific heat in J/(g·°C), cal/(g·°C), J/(kg·K) and more; temperature in °C, K or °F.
3. 3. Pick a substance if you want a table value. Choosing copper or water fills in its specific heat in the unit you selected; typing your own number switches back to a custom value.
4. 4. Read the working. Temperatures are converted to kelvin, so an impossible value at or below absolute zero is rejected, and every step is written out with your numbers.

The Specific Heat Equation

q = m × c × ΔT

ΔT = T₂ − T₁

q = m × L

q₁ + q₂ + q_cal = 0

q is the heat transferred, m the mass, c the specific heat capacity (the heat needed to raise 1 g of a substance by 1 °C) and ΔT the final minus the initial temperature. A positive q means heat flows in; a negative q means it flows out. A temperature difference is the same in °C and K, so ΔT does not need converting, but the calculator still converts each temperature to kelvin to catch impossible entries. One caution: this is a per-gram quantity. The molar heat capacity, in J/(mol·K), is a different number; divide by the molar mass (see the Molar Mass Calculator) to move between them.

Specific Heats Used

Values at 25 °C from OpenStax Chemistry 2e, Table 5.1, with silver, mercury, benzene, glycerin and glass from OpenStax College Physics 2e, Table 14.1. Real materials vary with temperature, purity and alloy, so treat these as textbook values.

Water's value, 4.184 J/(g·°C), is unusually high, which is why water warms and cools slowly and why it is the default liquid in calorimetry. It is also how the thermochemical calorie is defined: 1 cal = 4.184 J exactly.

Latent Heat of Fusion and Vaporization

During melting or boiling the temperature stays constant while heat breaks the bonds between particles, so q = mL replaces q = mcΔT. The table lists the values OpenStax College Physics 2e (Table 14.2) gives at the normal melting and boiling points and 1 atm.

Worked Examples

Heat to warm water (the default inputs). How much heat warms 250.0 g of water from 20.0 °C to 80.0 °C? q = 250.0 × 4.184 × 60.0 = 62 760 J, shown to three significant figures as 62.8 kJ.

Final temperature of a mixture. 250.0 g of copper at 310.0 °C goes into 400.0 g of water at 22.0 °C in an insulated cup. Heat lost by the copper equals heat gained by the water: T_f = (250.0 × 0.385 × 583.15 + 400.0 × 4.184 × 295.15) ÷ (250.0 × 0.385 + 400.0 × 4.184) = 310.8 K, or 37.7 °C. Adding a calorimeter that absorbs 25.0 J/°C lowers it to 37.4 °C.

Identifying a metal. 75.0 g of an unknown metal at 100.0 °C is dropped into 100.0 g of water at 22.0 °C, and the final temperature is 27.0 °C. Solving for c gives 0.382 J/(g·°C), within 5% of copper's 0.385.

Condensing steam. Condensing 10.0 g of steam at 100 °C releases q = −10.0 × 2256 = −22.6 kJ, which is why a steam burn is worse than a hot-water burn.

A full heating curve. Taking 135 g of ice at −15 °C to steam at 120 °C needs 4.238 kJ to warm the ice, 45.09 kJ to melt it, 56.48 kJ to warm the water, 304.6 kJ to boil it and 5.033 kJ to warm the steam: 415.4 kJ in total. Nearly three quarters of it goes into boiling.

Common Mistakes

• Skipping the phase change. q = mcΔT only works within one phase. If the sample melts or boils on the way, add an mL term for each, as in the heating curve.
• Mixing up J and kJ. Specific heats are in J/(g·°C); the answer is in joules unless you pick kJ. A factor of 1000 is the commonest slip.
• Getting the sign of ΔT wrong. ΔT is final minus initial. Cooling gives a negative ΔT and a negative q, and the calculator rejects a q whose sign contradicts the temperature change.
• Forgetting the calorimeter. In a real cup the container absorbs some heat. Enter its heat capacity, or accept that the final temperature will come out slightly high.
• Using the wrong specific heat for the phase. Ice, water and steam have three different values (2.093, 4.184 and 1.864 J/(g·°C)).

Frequently Asked Questions

What is the formula for specific heat?

c = q ÷ (m × ΔT), a rearrangement of q = mcΔT. Specific heat is the heat needed to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin). Its units are J/(g·°C), which equals J/(g·K).

What is the specific heat of water?

4.184 J/(g·°C) for liquid water at 25 °C, equal to 1.000 cal/(g·°C). Ice is 2.093 J/(g·°C) and steam 1.864 J/(g·°C), so a calculation that crosses a phase boundary must change the value.

What is the difference between specific heat and heat capacity?

Specific heat is per gram (or per kilogram), so it is a property of a substance. Heat capacity is for a whole object, C = mc, in J/°C. A calorimeter's heat capacity is the number you add to the water's m×c in a mixing problem.

How do I find the final temperature when two substances are mixed?

Set the heat lost by the hotter one equal to the heat gained by the cooler one: m₁c₁(T₁ − T_f) = m₂c₂(T_f − T₂). Solve for T_f. The answer always lies between the two starting temperatures, and the calculator warns you if your data say otherwise.

Why does the temperature stay constant while ice melts?

The heat goes into breaking the ordered bonds of the solid, not into faster motion, so the temperature holds at the melting point until all the solid has melted. The same happens at the boiling point. That heat is the latent heat, q = mL.

Is a calorie the same as a Calorie?

No. A calorie (cal) is 4.184 J. A food Calorie, written with a capital C, is a kilocalorie: 1000 cal or 4184 J. The unit menus use cal and kcal, so you can calculate in either and read the answer in the other.