In this set of practice questions, we will go over the main types of questions on calorimetry including the heat capacity, the heat of reaction, finding the final temperature of a mixture, constant pressure calorimetry, and constant-volume calorimetry.

A few important concepts and formulas you will need to solve these questions.

Heat capacity and specific heat are correlated by the following formula:

When solving a problem related to heat capacity and heat transfer, remember that most of the time, it is assumed the heat is not lost, and it only flows from the object with a higher temperature to the colder one:

The heat transfer to/from the calorimeter is determined by its temperature change and the heat capacity:

*q*_{cal }*= C*** _{cal}** x Δ

*T*

The heat capacity of the calorimeter is determined experimentally and is already known when measuring the heat/enthalpy of a reaction.

Remember, the enthalpy change **(****Δ***H)*** is equal to the heat** when the **pressure is constant**:

On the other hand, a** bomb calorimeter** is an equipment that measures **Δ***E ***for combustion reactions**. This can be seen in the equation of internal energy change:

Δ*E = q + w*

* *

w = *P*Δ*V* and Δ*V* is zero when the volume is constant. So, for a constant volume, **Δ***E*** = q**, and therefore, the

**bomb calorimeter measures the energy change**of a chemical reaction.

The experiment is carried out in an insulated sealed vessel called a *bomb *which is designed to withstand high pressures. The bomb is placed in a water container and by measuring its temperature change caused by the reaction, we determine the heat of the calorimeter.

The links to the corresponding topics are given below:

- Energy Related to Heat and Work
- Endothermic and Exothermic Processes
- Heat Capacity and Specific Heat
- Heat Capacity Practice Problems
- What is Enthalpy
- Constant-Pressure Calorimetry
- Bomb calorimeter – Constant Volume Calorimetry
- Stoichiometry and Enthalpy of Chemical Reactions
- Hess’s Law and Enthalpy of Reaction
- Hess’s Law Practice Problems
- Standard Enthalpies of Formation
- Enthalpy of Reaction from Enthalpies of Formation

#### Practice

How much heat in kJ is required to warm 1.50 L of water from 25.0 ^{o}C to 100.0 °C? (Assume a density of 1.0 g/mL for the water.)

4.70 x 10^{2} kJ

The water doesn’t go through any phase changes so, we can use the following formula for the relationship between the amount of heat needed to raise the temperature of a given substance and the corresponding temperature increase:

** **q = m × C_{s} × ΔT

Where, q is the amount of heat in J, m is the mass, C_{s} is the specific heat capacity of the material in J/g ^{o}C , and ΔT is the change in temperature (T_{final} – T_{initial}). Remember, the specific heat is the quantity of heat energy necessary to increase the temperature of 1 g of a substance by 1°C and it is 4.18 J/g ^{o}C for water. Therefore, to find the amount of heat for a given mass, we multiply it by this mass.

Before using the numbers, make sure the units match. The C_{s }is given for a gram of water so, we need to convert the volume to mL first, and then to mass in g.

\[{\rm{V}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1,000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{mL}}\]

m = d x V

\[{\rm{m}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{g}}}}{{{\rm{mL}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{g}}\]

And now, we can plug the numbers into the equation of the heat:

q = m × C_{s} × ΔT

\[{\rm{q}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}{\rm{.18}}\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(100}}{\rm{.0}}\;{\rm{ – }}\;{\rm{25}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.70}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{{\rm{5}}\;}}{\rm{J}}\]

To convert the J to kJ, we need to divide it by 1,000, so it is 4.70 x 10^{2} kJ

What is the final temperature when a 40 g sample of water at 90 °C is mixed with a 60 g sample at 25 °C?

T_{f }= 51 °C

The basis of solving this problem is the assumption that **q _{lost} = q_{gained}**

_{.}The heat flows from the warmer sample (q

_{lost}) to the cooler one (q

_{gained}), and the amount of heat lost by the warmer sample is equal to the amount of heat gained by the cooler sample. This means that there is no heat lost to the surroundings.

Recall that q = m · C · ΔT, where m is the mass, C is the specific heat, and ΔT is the temperature change.

We can also write this as q_{lost} + q_{gained }= 0 or

q_{1 }+ q_{2} = 0

For our problem, q_{1 }= mCΔT_{1 }, q_{2 }= mCΔT_{2 }

so,** **mCΔT_{1 }+** **mCΔT_{2} = 0

ΔT_{1 }= T_{f }– 90, and ΔT_{2 }= T_{f }– 25

Where T_{f }is the final temperature of the system, so it is the same for the worm and cold samples.

Plug in the numbers and solve for T_{f }in the following equation:

40 g x C x (T_{f }– 90) + 60 g x C x (T_{f }– 25) = 0

Because C (heat capacity of water) appears on both sides of the equation and can be canceled out.

40 g x (T_{f }– 90) + 60 g x (T_{f }– 25) = 0

T_{f }= 51 °C

Notice that the equation can also be set up by using kelvin (K).

How much heat does it take to increase the temperature of a 540.6-g sample of Fe from 20.0 °C to 84.3 °C? The specific heat of iron = 0.450 J/g °C.

15.6 kJ

A similar question to that in problem 8, and we are going to use the following equation:

** **q = m × C_{s} × ΔT

Where, q is the amount of heat in J, m is the mass, C_{s} is the specific heat capacity of the material in J/g ^{o}C , and ΔT is the change in temperature (T_{final} – T_{initial}). Remember, the specific heat is the quantity of heat energy necessary to increase the temperature of 1 g of a substance by 1°C.

All we need to do is to plug the numbers into the equation of the heat:

\[{\rm{q}}\;{\rm{ = }}\;{\rm{540}}{\rm{.6}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{0}}{\rm{.450}}\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(84}}{\rm{.3}}\;{\rm{ – }}\;{\rm{20}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{15}}{\rm{.6}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{{\rm{3}}\;}}{\rm{J}}\]

To convert the J to kJ, we need to divide it by 1,000, so it is 15.6 kJ

Calculate the specific heat capacity of a metal if a 17.0 g sample requires 481 J to change the temperature of the metal from 25.0 °C to 67.0 °C?

0.674 J/g°C

We are going to rearrange the following equation for the relation of heat, specific heat capacity, and temperature change:

** **q = m × C × ΔT

\[{\rm{C}}\;{\rm{ = }}\;\frac{{\rm{q}}}{{{\rm{m}}\;{\rm{\Delta T}}}}\]

\[{\rm{C}}\;{\rm{ = }}\;\frac{{{\rm{481}}\;{\rm{J}}}}{{{\rm{17}}{\rm{.0}}\;{\rm{g}}\;{\rm{(67}}{\rm{.0}}\;{\rm{ – }}\;{\rm{25}}{\rm{.0)}}{\;^{\rm{o}}}{\rm{C}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.674}}\;{\rm{J}}}}{{{\rm{g}}{\;^{\rm{o}}}{\rm{C}}}}\;\]

Calculate the energy of combustion for one mole of butane if burning a 0.367 g sample of butane (C_{4}H_{10}) has increased the temperature of a bomb calorimeter by 7.73 °C. The heat capacity of the bomb calorimeter is 2.36 kJ/ °C.

2.88 x 10^{3} kJ

Because the combustion of butane increased the temperature of a bomb calorimeter, it is an exothermic reaction. This means, the heat goes from the reaction to the calorimeter and assuming there is no heat loss, the heat released by the reaction is equal to the heat absorbed by the calorimeter:

-q_{reaction} = q_{cal}

Since the reaction occurs under conditions of constant volume, q_{rxn} = ΔE_{rxn, }so we need to calculate the q_{cal }using the following formula:

q_{cal} = C_{cal} x ΔT

q_{cal} = 2.36 kJ/ °C x 7.73 °C = 18.2 kJ

Therefore, q_{rxn }= -18.2 kJ

This is the change in the internal energy of the reaction for that specific amount of butane that was burned.

To get ΔE_{rxn} per mole of butane, we need to divide q_{rxn} by the number of moles that actually reacted. To find the number of moles, we use the mas and molar mass of butane:

n(C_{4}H_{10}) = 0.367 g/58.0 g/mol = 0.00633 mol

And after this, we divide the heat of the reaction by the number of moles of butane, to get the heat of the reaction per mole of butane:

ΔE_{rxn }= 18.2 kJ/0.00633 mol = 2.88 x 10^{3} kJ/mol

Alternatively, we can set up a cross-multiplication correlation:

0.00633 mol C_{4}H_{10 }– 18.2 kJ

1 mol C_{4}H_{10 }– X kJ

X = 2.88 x 10^{3} kJ/mol

How many joules of energy is required to melt 40.0 g of ice at 0 °C? The heat of fusion (Δ*H*_{fus}) for ice is 334.0 J/g.

1.34 x 10^{3} J

Since the ice/water is already at 0 ^{o}C (melting/freezing temperature), the added heat does not change the temperature of the ice and water mixture, but rather it is used for the transition from solid to liquid.

To calculate the amount of energy required to achieve a state change, in this case melting, of the substance that is already at the state-change temperature, we use the following formula:

q = mΔ*H*_{fus}

where q is the quantity of heat energy, m is the mass, and Δ*H*_{fus} is the enthalpy of fusion, sometimes called the heat of fusion. It is the energy required to fuse (melt) a gram (or a mole) of a substance.

Plugging the numbers in the formula, we get:

q = 40.0 g x 334.0 J/g = 13,360 J

Rounding off to three significant figures, we have q = 1.34 x 10^{3} J

How many kJ of energy does it take to change 36.0 g of ice at -15.0 °C to water at 0. °C ? The specific heat of ice is 2.10 J/g°C and the heat of fusion (Δ*H*_{fus}) for ice is 334.0 J/g. Ignore the significant figures for this problem.

1.32 x 10^{4} J

We mentioned in the previous problem, that since the ice/water is already at 0 ^{o}C (melting/freezing temperature), the added heat does not change the temperature of the ice and water mixture, but rather it is used for the transition from solid to liquid. This is not the case here because the ice is at -15.0 oC and before melting, it needs to be first warmed to 0 ^{o}C. Therefore, there are two stages in this process; 1) heating the ice to 0 ^{o}C, 2) melting the ice.

The overall heat of this process then is:

q = q_{heating} + q_{melting}

q_{heating }is calculated by the formula we have already used several times:

q_{heating }= m × C × ΔT

\[{{\rm{q}}_{{\rm{heating}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{2.10\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(0}}\;{\rm{ – }}\;( – {\rm{15}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{1,134}}\;{\rm{J}}\]

To calculate the amount of energy required to melt the ice, we use the following formula:

q = mΔ*H*_{fus}

\[{{\rm{q}}_{{\rm{melting}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{334.0\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;}}\;{\rm{ = }}\;{\rm{12,024}}\;{\rm{J}}\]

q = 1134 J + 12,024 J = 13,158 J

Rounding of to three significant figures, we have q = 1.32 x 10^{4} J or 13.2 kJ

The enthalpy change for the reaction is given below:

2CH_{3}OH(*l*) + 3O_{2}(*g*) → 4H_{2}O(*l*) + 2CO_{2}(*g*) Δ*H =* -1452.8 kJ

**a)** What quantity of heat is released for each mole of water formed?

**b) **What quantity of heat is released for each mole of oxygen reacted?

**a)** 363.2 kJ

**b)** 484.27 kJ

**a)** The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation**. Because the coefficient in front of H_{2}O is a 4, the formation of one mole of water will release:

-1452.8 kJ ÷ 4 = 363.2 kJ

**b)** According to the equation, when 3 moles of oxygen react, 8 kJ energy is released. Therefore, for the reaction of one-mole oxygen,

Δ*H = *-1452.8 kJ ÷ 3 = 484.27 kJ

How much heat will be released if 44.8 g of SO_{2} is reacted with an excess of oxygen according to the following chemical equation?

2SO2(*g*) + O_{2}(*g*) → 2SO_{3}(*g*), Δ*H*° = –198 kJ

69.2 kJ

The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation**. So, in this reaction, the combustion of two moles of SO_{2 }produces 198 kJ of heat. What we need to do here is first convert the mass of SO_{2 }to moles, and then calculate the amount of heat that will be released from this amount considering that burning two moles of SO_{2 }gives 198 kJ of heat.

n(SO_{2}) = 44.8 g/64.1 g/mol = 0.699 mol

Let’s now set up a cross multiplication, which would read as “2 mol SO_{2 }gives 198 kJ heat, 0.699 mol SO_{2 }will give an unknown amount of heat:

2 mol SO_{2 }– 198 kJ

0.699 mol SO_{2 }– X kJ

X = 69.2 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{198}}\;{\rm{kJ}}}}{{\cancel{{{\rm{2}}\;{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.699 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;{\rm{69}}{\rm{.2}}\;{\rm{kJ}}\]

So, 69.2 kJ heat will be released if 44.8 g (0.699 mol ) of SO_{2 }is reacted according to the given chemical equation.

What is Δ*H*° for the following reaction

2C_{6}H_{6}(*l*) + 15O_{2}(*g*) → 12CO_{2}(*g*) + 6H_{2}O(*l*), Δ*H*° = ? kJ

if the consumption of 27.3 g of benzene (C_{6}H_{6}) produces 1144 kJ of heat?

6.54 x 10^{3} kJ

The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation**. So, in this reaction, we need to determine the combustion of two moles of C_{6}H_{6.}

Convert the mass of C_{6}H_{6}_{ }to moles, and then calculate the amount of heat that will be released from 2 moles of C_{6}H_{6} considering that burning 27.3 g C_{6}H_{6 }gives 1144 kJ of heat.

n(C_{6}H_{6}) = 27.3 g/78.1 g/mol = 0.350 mol

Let’s now set up a cross multiplication, which would read as “0.350 mol C_{6}H_{6}_{ }gives 1144 kJ heat, 2 mol C_{6}H_{6}_{ }will give an unknown amount of heat:

0.350 mol C_{6}H_{6}_{ }– 1144 kJ

2 mol C_{6}H_{6}_{ }– X kJ

X = 6.54 x 10^{3} kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{1144}}\;{\rm{kJ}}}}{{\cancel{{{\rm{0}}{\rm{.350}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}}}}}\;{\rm{ \times }}\;{\rm{2}}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.54}}\;{\rm{ \times }}\;{10^3}\;{\rm{kJ}}\]

Based on the heat of reaction for the chlorination of methane, how much heat will be released if 233.6 grams of hydrochloric acid are formed?

CH_{4}(*g*) + 3Cl_{2}(*g*) → CHCl_{3}(*l*) + 3HCl(*g*), Δ*H*° = -334 kJ

713 kJ

The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation **which imply their mole numbers. Therefore, converting the mass to moles is always going to be a good idea.

n(HCl) = 233.6/36.5 g/mol = 6.40 mol

According to the chemical equation, for every three moles of HCl forming, there is 334 kJ heat released. So, we need to find how much heat will be released if 6.40 mol HCl is formed.

We can set up a cross multiplication, which would read as “3 mol HCl_{ }gives 334 kJ heat, 6.40 mol HCl_{ }will give an unknown amount of heat:

3 mol HCl_{ }– 334 kJ

6.40 mol HCl_{ }– X kJ

X = 713 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{334}}\;{\rm{kJ}}}}{{{\rm{3 }}\cancel{{{\rm{mol HCl}}}}}}\;{\rm{ \times }}\;{\rm{6}}{\rm{.40}}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ = }}\;713\;{\rm{kJ}}\]

Calculate how many kJ of heat-energy will be released when 12.65 g of magnesium carbonate reacts with 650. mL of 0.400* M* hydrochloric acid?

MgCO_{3}(*s*)* *+* *2HCl(*aq*)* *→* *MgCl_{2}(*aq*)* *+* *H_{2}O(*l*)* *+* *CO_{2}(*g*), Δ*H*°* *=* *–112* *kJ

14.6 kJ

First, let’s convert all the quantities to moles.

n(MgCO_{3}) = 12.65 g/84.3 g/mol = 0.150 mol

The moles of HCl are calculated using the equation for molarity:

M = n/V

n = MV

n(HCl) = MV = 0.400 M x 0.650 L = 0.260 mol

We have the moles of both reactants, and therefore, we need to find the limiting reactant in order to do the calculations for the heat of the reaction. Remember, the limiting reactant is the one that gives less product. So, to find the LR, we determine whether the MgCO_{3}_{ }or HCl could produce less product based on their moles and the reaction stoichiometry. Let’s do the calculations based on the number of MgCl_{2} moles formed:

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}\;{\rm{from}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}} \right){\rm{\; = }}\;{\rm{0}}{\rm{.150}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.150}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}\;{\rm{from}}\;{\rm{HCl}}} \right){\rm{\; = }}\;{\rm{0}}{\rm{.260}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.130}}\;{\rm{mol}}\]

So, HCl gives less MgCl_{2}, therefore, it is the limiting reactant, and we need to calculate the amoiunt of heat beased on 0.260 mol HCl.

We can set up a cross multiplication, which would read as “2 mol HCl_{ }gives 112 kJ heat, 0.260 mol HCl_{ }will give an unknown amount of heat:

2 mol HCl_{ }– 112 kJ

0.260 mol HCl_{ }– X kJ

X = 14.6 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{112}}\;{\rm{kJ}}}}{{{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.260}}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ = }}\;14.6\;{\rm{kJ}}\]