The calorimetry calculator can help you solve complex calorimetry problems. It can analyze the heat exchange between up to 3 objects. Additionally, it can find the enthalpy change of a chemical reaction inside a coffee-cup calorimeter.

Read on to learn what calorimetry is and how to solve calorimetry problems with the proper equations.

What is calorimetry?

Calorimetry is a science where you try to find the heat transfer during a chemical reaction, phase transition, or temperature change. Calorimetry experiments are based on the law of conservation of energy. This law states that the total energy of an isolated system is constant.

In physics, heat is also known as thermal energy. So, if there is an isolated system where objects exchange heat, the total heat change equals 0.

🙋 Learn more about thermal energy in the thermal energy calculator.

0=ΔQ1+ΔQ2+...+ΔQi\scriptsize 0 = \Delta Q_{1} + \Delta Q_{2} + ... + \Delta Q_{i}

Objects tend towards thermal equilibrium. So, two objects of different temperatures will exchange heat on contact. The object with a higher temperature will give away heat, while the other will absorb it. After some time, they will have the same temperature.

Calorimetry equation

It's not easy to measure the heat transfer directly. That's why we use constant pressure calorimeters – containers that provide constant pressure and thermal isolation from the surroundings. In a calorimeter, we can measure temperature change with a thermometer. This equation binds temperature change and heat:

ΔQ=mcΔT\scriptsize \Delta Q = m \cdot c \cdot \Delta T

where:

  • ΔQ\small \Delta Q – Heat change;
  • m\small m – Mass of an object;
  • c\small cHeat capacity of an object (the amount of heat it needs to change the temperature by exactly 1 °C or 1 K); and
  • ΔT\small \Delta T – Temperature change, and its equal to the difference of final and initial temperature: ΔT=TfinalTinitial\small \Delta T = T_{\text{final}} - T_{\text{initial}} .

When the temperature change is positive (the temperature of an object has risen), the heat change is also positive. Then, we say that an endothermic process has occurred. Endothermic processes absorb thermal heat to occur (i.e., ice melting).

A negative temperature and heat change means an exothermic reaction occurred. In the process, the heat is lost, and the temperature drops.

An object at a phase transition temperature absorbs or releases heat until the phase change happens without changing the temperature. Then, the heat change is equal to mass times latent heat of a substance:

ΔQ=mH\scriptsize \Delta Q = m \cdot H

where:

The heat of transition is positive when an object changes phase from one of higher order to one of lower order (i.e., ice to water). In the opposite situation, it's negative (i.e., steam to water).

So, for example, if you need to find the change of heat from ice at -40 °C to water at 20 °C, you need to calculate three different heat absorptions:

  1. First, the ice absorbs heat until it reaches 0 degrees.
  2. Then, it absorbs heat until all of it melts.
  3. Lastly, it absorbs heat until it reaches 20 °C. Here, the object that changes heat is water, so you need to use water's heat capacity.

We will explain these calculations with an example below.

How to solve calorimetry problems?

To solve calorimetry problems, you need to follow a simple instruction:

  1. First, write down everything you know about each object and what you need to find.

  2. Second, write an equation for heat change for each object.
    ΔQ = m × c × (Final temperature - Initial temperature)

  3. Then, write the equation for total heat change in the system.

    0 = ΔQ1 + ΔQ2 + ... + ΔQi

    0 = m1 × c1 × (T final - T init1) + m2 × c2 × (T final - T init2) + ... + mi × ci × (T final - T initI)

  4. Transform the equation so that the unknown is on the left side and everything else is on the right side.

  5. Input all known values into the equation and find the unknown.

Don't worry if you don't know how to use the instructions yet. Now, we will look at some examples, so you know how to solve calorimetry problems easily.

Calorimetry problems – example 1

You poured 1.4 kg\small 1.4 \text{ kg} of mercury into a calorimeter (0.5 kg\small 0.5 \text{ kg}) containing 3 kg\small 3 \text{ kg} of water at 20°C\small 20 \degree \text{C}. The mercury had a temperature of 100°C\small 100 \degree \text{C}. The temperature of water in the calorimeter increased to 21.21°C\small 21.21 \degree \text{C}. Calculate the specific heat capacity of mercury. The calorimeter constant is 0.0922 cal/gK\small 0.0922\ \text{cal}/\text g \cdot \text K, and water's heat capacity is 0.999 cal/gK\small 0.999\ \text{cal}/\text g \cdot \text K.

1. Following the instructions above, we should first write out everything we know about the system.

Calorimeter (copper)

Water

Mercury

Mass [kg]\small \text{[kg]}

0.5

3

1.4

Specific heat capacity [cal/gK]\small [\text{cal}/\text g \cdot \text K]

0.092

0.999

?

Initial temperature [°C]\small \text{[} \degree \text{C]}

20

20

100

Final temperature [°C]\small \text{[} \degree \text{C]}

21.21

21.21

21.21

2. Now, write the heat change equation for each object:

  • Calorimeter:
ΔQc=mccc(TFTI(c))\qquad \scriptsize \Delta Q_{c} = m_{c} \cdot c_{c} \cdot (T_{F} - T_{I(c)})
  • Water:
ΔQw=mwcw(TFTI(w))\qquad \scriptsize \Delta Q_{w} = m_{w} \cdot c_{w} \cdot (T_{F} - T_{I(w)})
  • Mercury:
ΔQm=mmcm(TFTI(m))\qquad \scriptsize \Delta Q_{m} = m_{m} \cdot c_{m} \cdot (T_{F} - T_{I(m)})

3. Write down the calorimetry equation.

0=ΔQc+ΔQw+ΔQm\scriptsize 0 = \Delta Q_{c} + \Delta Q_{w} + \Delta Q_{m}
0=  mccc(TFTI(c)) +mwcw(TFTI(w)) +mmcm(TFTI(m))\scriptsize \begin{split} 0= \ \ &m_{c} \cdot c_{c} \cdot (T_{F} - T_{I(c)})\ + \\ &m_{w} \cdot c_{w} \cdot (T_{F} - T_{I(w)})\ + \\ &m_{m} \cdot c_{m} \cdot (T_{F} - T_{I(m)}) \end{split}

4. Transform the equation so that the unknown is on the left side and everything else is on the right side.

 ⁣mmcm(TFTI(m))=mccc(TFTI(c))+mwcw(TFTI(w))\! \scriptsize - m_{m} \cdot c_{m} \cdot (T_{F} - T_{I(m)}) = \\ \quad \! m_{c} \!\cdot\!c_{c}\! \cdot\! (T_{F}\! -\! T_{I(c)}\!)\! +\! m_{w}\! \cdot\! c_{w}\! \cdot\! (T_{F}\! -\! T_{I(w)}\!)