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Calorimetry Calculator

Table of contents

What is calorimetry?Calorimetry equationHow to solve calorimetry problems?Calorimetry problems – example 1Calorimetry problems — example 2Chemical reaction in a coffee-cup calorimeterHow to use the calorimetry calculator?FAQs

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


  • 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


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)



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




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




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




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




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)}\!)
 ⁣cm=mccc(TFTI(c))+mwcw(TFTI(w))mm(TFTI(m))\! \scriptsize c_{m}\! = \!- \frac{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 (T_{F} - T_{I(m)})}

5. Lastly, input all known values into the equation and find the unknown:

 ⁣cm=5000.092(21.2120) +30000.999(21.2120)/[1400(21.21100)]=55.66+3626.37110306=3682.03110306=0.0333cal/gK\!\scriptsize \begin{split} c_{m} \! &= \!- 500 \!\cdot \! 0.092 \!\cdot \!(21.21\! - \!20)\ +\\ \! &\qquad 3000 \!\cdot\! 0.999\! \cdot \!(21.21 \!- \!20)/ \\ &\qquad[1400 \cdot (21.21 - 100)] \\[1em] &=- \frac{55.66+ 3626.37}{- 110306} \\[1em] &= \frac{3682.03}{110306} \\[1em] &=0.0333 \thinspace \text{cal}/\text g \cdot \text K \end{split}

The specific heat capacity of mercury is 0.0333 cal/gK\small 0.0333\ \text{cal}/\text g \cdot \text K.

💡 You can use either degrees Celsius (°C) or kelvin (K) to calculate temperature change. But when you need to input just temperature (not temperature change) into the equation, always use kelvin! Be even more careful with other variables. For example, if you want to use the heat capacity in J/(kgK)\small \rm J/(kg \cdot K) , you need to have mass in kilograms (kg\small \rm kg). Then, you will get the heat change in Joules (J\small J). But if you have a specific heat capacity in cal/(gK)\small \rm cal/(g \cdot K), the mass should be in grams (g\small \rm g), and the heat change will be in calories (cal\small \rm cal).

Calorimetry problems — example 2

Example 2. You dropped 200 g\small 200 \text{ g} of ice into a bucket containing 2 kg\small 2 \text{ kg} of water at 66 °C\small 66\ \degree \text{C}. Let's assume the only objects exchanging heat are water and ice. The temperature of ice was 25 °C\small -25\ \degree \text{C}. Find the final temperature of the system. The heat of fusion of ice is 334 J/g\small 334\ \text{J}/\text g. The specific heat capacity of water is 4.18 J/gK\small 4.18\ \text{J}/\text g \cdot \text K, and of ice, it's 2.05 J/gK\small 2.05\ \text{J}/\text g \cdot \text K.

1. Like before, first write down everything you know about the system:



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



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



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



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



2. Write the heat change equation for each object:

  • Ice:
ΔQ1=mici(TfusionTI(i))ΔQ2=miΔHfusionΔQ3=micw(TFTfusion)\quad \scriptsize \begin{align*} \Delta Q_{1} &= m_{i} \cdot c_{i} \cdot (T_{\text{fusion}} - T_{I(i)}) \\[.5em] \quad \Delta Q_{2} &= m_{i} \cdot \Delta H_{\text{fusion}} \\[.5em] \quad \Delta Q_{3} &= m_{i} \cdot c_{w} \cdot (T_{F} - T_{\text{fusion}}) \end{align*}
  • Water:
ΔQw=mwcw(TFTI(w))\qquad \scriptsize \Delta Q_{w} = m_{w} \cdot c_{w} \cdot (T_{F} - T_{I(w)})

3. Write down the calorimetry equation:

0=ΔQ1+ΔQ2+ΔQ3+ΔQw\scriptsize 0 = \Delta Q_{1} + \Delta Q_{2} + \Delta Q_{3} + \Delta Q_{w}
0=  mici(TfusionTI(i)) +miΔHfusion +micw(TFTfusion) +mwcw(TFTI(w))\scriptsize \begin{split} 0=\ \ &m_{i} \cdot c_{i} \cdot (T_{\text{fusion}} - T_{I(i)})\ + \\ &m_{i} \cdot \Delta H_{\text{fusion}}\ + \\ &m_{i} \cdot c_{w} \cdot (T_{F} - T_{\text{fusion}})\ + \\ &m_{w} \cdot c_{w} \cdot (T_{F} - T_{I(w)}) \end{split}

4. Transform the equation:

  • Step-by-step transformation:
 ⁣micw(TFTfusion)mwcw(TFTI(w))=mici(TfusionTI(i))+miΔHfusion\!\scriptsize \! -m_{i} \!\cdot \! c_{w} \!\cdot \!(T_{F}\! -\! T_{\text{fusion}}\!)\! -\! m_{w}\! \cdot\! c_{w}\! \cdot \!(T_{F}\! - \!T_{I(w)}\!) \! \\ \quad = \!m_{i} \!\cdot \! c_{i} \!\cdot \!(T_{\text{fusion}}\! -\! T_{I(i)}\!) \!+ \!m_{i} \!\cdot\! \Delta H_{\text{fusion}}

 ⁣micwTF+micwTfusionmwcwTF+mwcwTI(w)=mici(TfusionTI(i))+miΔHfusion\!\scriptsize \! -m_{i}\! \cdot \! c_{w} \!\cdot \!T_{F} \!+ \!m_{i}\! \cdot \! c_{w}\!\cdot\! T_{\text{fusion}}\! -\! m_{w}\! \cdot \! c_{w}\! \cdot \!T_{F}\! +\\ \quad \!m_{w}\! \cdot \! c_{w} \!\cdot \!T_{I(w)}\! \\ \quad =\! m_{i} \!\cdot \! c_{i}\! \cdot \!(T_{\text{fusion}}\! -\! T_{I(i)}\!) \!+ \!m_{i} \!\cdot \!\Delta H_{\text{fusion}}
 ⁣micwTFmwcwTF=mici(TfusionTI(i))+miΔHfusionmicwTfusionmwcwTI(w)\!\scriptsize \!-m_{i} \!\cdot \! c_{w}\! \cdot \!T_{F}\! -\! m_{w}\! \cdot \! c_{w} \!\cdot \!T_{F} \! \\ \quad = \!m_{i}\! \cdot \! c_{i} \!\cdot\! (T_{\text{fusion}}\! -\! T_{I(i)}\!) \!+ \!m_{i}\! \cdot \!\Delta H_{\text{fusion}}\! -\\\qquad \!m_{i}\! \cdot \! c_{w}\!\cdot \!T_{\text{fusion}}\! -\! m_{w}\! \cdot \!c_{w}\! \cdot\! T_{I(w)}
(micwmwcw)TF=mici(TfusionTI(i))+miΔHfusionmicwTfusionmwcwTI(w)\scriptsize \!(\!-m_{i} \!\cdot \! c_{w} \! - \!m_{w} \!\cdot \! c_{w})\! \cdot \!T_{F}\! \\ \quad = \!m_{i} \!\cdot \! c_{i} \!\cdot \!(T_{\text{fusion}} \!- \!T_{I(i)}\!) \!+ \!m_{i} \!\cdot \!\Delta H_{\text{fusion}} \!-\!\\\qquad m_{i} \!\cdot \! c_{w}\!\cdot\! T_{\text{fusion}} \!-\! m_{w}\! \cdot \! c_{w} \!\cdot\! T_{I(w)}
  • Final equation:
 ⁣TF=mici(TfusionTI(i))+miΔHfusionmicwTfusionmwcwTI(w)/[micwmwcw]\! \scriptsize T_{F} \! = \!m_{i} \!\cdot \! c_{i}\! \cdot \!(T_{\text{fusion}} \!- \!T_{I(i)}\!)\! + \!m_{i} \!\cdot \!\Delta H_{\text{fusion}} \!-\!\\\qquad m_{i}\! \cdot \! c_{w}\!\cdot\! T_{\text{fusion}}\! - \!m_{w} \!\cdot \! c_{w} \!\cdot \!T_{I(w)}\\\qquad/[-m_{i} \cdot c_{w} - m_{w} \cdot c_{w}]

Ugh, that's a tough one!

5. Input all the know values (Remember to use correct units!) and find the final temperature:

TF=2002.05(0(25))+2003342004.18273.1520004.18339.15/[2004.1820004.18]=324.77 K=51.62 °C\scriptsize \begin{split} T_{F} &= \! 200 \!\cdot \!2.05 \!\cdot \!(0 \!- \!(\!-25)\!)\! + \!200 \!\cdot \!334 \!-\\& \quad 200 \!\cdot\! 4.18 \!\cdot\! 273.15 \!-\! 2000\! \cdot \! 4.18\! \cdot\! 339.15\\ &\quad/[-200 \cdot 4.18 - 2000 \cdot 4.18] \\[.5em] &= 324.77\ \text{K} \\ &= 51.62\ \degree \text{C} \end{split}

In our calorimetry calculator, input the material that changes phase (in this case – ice) as object 1.

Chemical reaction in a coffee-cup calorimeter

A coffee-cup calorimeter is a simple version of a constant-pressure calorimeter. It's usually made of styrofoam. Using this type of calorimeter, you can measure enthalpy change during chemical reactions. To do that, first, you need to find the heat change of a calorimeter during the chemical reaction. The heat change of a solution inside the calorimeter is equal, so:

ΔQsolution=ΔQcalorimeter\footnotesize \Delta Q_{\text {solution}} = - \Delta Q_{\text {calorimeter}}

With this value, you can easily say if a reaction was exo- or endothermic. To find the enthalpy change, all you have to do is use this equation:

ΔH=molar massΔQsolutionmasssubstance\footnotesize \Delta H = \text {molar mass} \cdot \frac{\Delta Q_{\text {solution}}}{\text {mass}_{\text {substance}}}

🙋 To learn more about enthalpy, go to enthalpy calculator.

How to use the calorimetry calculator?

To use our calorimetry calculator, first, choose what you want to research:

  1. A chemical reaction in a coffee cup calorimeter:

    • Input everything you know about a substance.

    • If you want to find enthalpy change, input the mass of a substance, and its molar mass.

  2. Heat exchange between several objects:

    • Choose the number of objects in an isolated system. You can choose between 2 or 3.

    • Type in what you know about each object. Set the correct units.

    • If the object changes phase during the experiment, input it as object 1.

    • The unknown value will appear in seconds.

Thank you for using our calorimetry calculator! We hope you learned what calorimetry is and how to solve calorimetry problems! We have many other thermodynamics calculators, so feel free to check them out!


How to identify unknown metal using calorimetry?

To identify an unknown metal with calorimetry, you need to find its specific heat capacity and compare its value with an appropriate source.

Where is calorimetry used in industry?

Calorimetry helps food and drug research and is commonly used in science laboratories or coal and fuel industries. They are also used in iron, steel, and cement plants.

How to find calorimeter constant?

To find the calorimeter constant, you can:

  • Check the specific heat capacity of the material it is made of; or

  • Carry an experiment, and use the calorimetry equation:

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

I want to research...

Cofee-cup calorimeter experiment

Enthalpy change:

Lookup specific heat capacity

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