Calculate heat of dissolution, solvent mass, specific heat, temperature change, or moles of solute using the other four entered values.
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Heat Of Dissolution Formula
The heat of dissolution is calculated from the heat gained or lost by the solvent, divided by the moles of solute. The calculator uses the calorimetry relationship with a negative sign because the heat change of dissolving is opposite the heat change measured in the solvent.
\Delta H_{diss} = -\frac{m c \Delta T}{n}- ΔHdiss = heat of dissolution, usually in J/mol or kJ/mol
- m = mass of solvent
- c = specific heat of the solvent
- ΔT = change in temperature of the solution or solvent
- n = moles of solute dissolved
If you leave the heat of dissolution field blank, the calculator uses the main formula above. If you leave another field blank, it rearranges the same formula:
m = -\frac{\Delta H_{diss} n}{c \Delta T}c = -\frac{\Delta H_{diss} n}{m \Delta T}\Delta T = -\frac{\Delta H_{diss} n}{m c}n = -\frac{m c \Delta T}{\Delta H_{diss}}The calculator converts entries to base units before solving: J/mol for heat of dissolution, grams for mass, J/g°C for specific heat, °C for temperature change, and mol for amount of solute. A positive temperature change gives a negative heat of dissolution, which means the dissolving process released heat. A negative temperature change gives a positive heat of dissolution, which means the dissolving process absorbed heat.
Common Solvent Specific Heat Values
Use the correct specific heat for the solvent when it is known. For dilute aqueous solutions, water is often used as an approximation.
| Solvent | Specific Heat | Notes |
|---|---|---|
| Water | 4.184 J/g°C | Common value for aqueous calorimetry problems |
| Ethanol | 2.44 J/g°C | Lower heat capacity than water |
| Methanol | 2.53 J/g°C | Typical liquid value near room temperature |
| Acetone | 2.15 J/g°C | Use only if acetone is the solvent |
Interpreting Heat Of Dissolution Results
| Result Sign | Temperature Change | Meaning |
|---|---|---|
| Negative ΔHdiss | Temperature increases | Exothermic dissolution, heat is released |
| Positive ΔHdiss | Temperature decreases | Endothermic dissolution, heat is absorbed |
| Near zero | Very small change | Small heat effect or limited measurement precision |
Example Problems
Example 1: Calculate heat of dissolution
You dissolve 0.050 mol of solute in 100 g of water. The specific heat is 4.184 J/g°C, and the temperature increases by 6.0°C.
\Delta H_{diss} = -\frac{100 \times 4.184 \times 6.0}{0.050}The heat of dissolution is:
ΔHdiss = -50,208 J/mol = -50.208 kJ/mol
Example 2: Calculate moles of solute
A dissolution has ΔHdiss = 25,000 J/mol. The solvent mass is 150 g, the specific heat is 4.184 J/g°C, and the temperature change is -2.0°C.
n = -\frac{150 \times 4.184 \times (-2.0)}{25000}The amount of solute is:
n = 0.050208 mol
FAQ
Why is there a negative sign in the heat of dissolution formula?
The negative sign accounts for heat flow direction. The solvent temperature change shows the heat gained or lost by the solvent. The dissolving process has the opposite heat change. If the solvent warms up, the dissolving process released heat, so ΔHdiss is negative. If the solvent cools down, the dissolving process absorbed heat, so ΔHdiss is positive.
Should temperature change be entered in Celsius or Kelvin?
For a temperature difference, 1°C and 1 K have the same size. A change of 5°C is the same as a change of 5 K. If you enter Fahrenheit, the calculator converts the temperature difference using 5/9.
Can this be used for any solvent?
Yes, if you enter the correct solvent mass and specific heat. For water-based solutions, 4.184 J/g°C is commonly used as an approximation. For other solvents, use the specific heat of that solvent, because the calculated heat of dissolution depends directly on that value.