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Wize University Chemistry Textbook > Thermodynamics: Part 1: (0th law, 1st law, Calorimetry, Enthalpy)

Calorimetry

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Heat Capacity and Calorimetry

  • Heat capacity (C): has units of J/oK or J/oC and describes the amount of heat needed to heat up the substance or object by one degree
  • Specific heat capacity (c): has units of J/goK or J/goC and described the amount of heat required to heat up 1g of a substance by one degree.
  • We also have a molar heat capacity (units J/Kmol, or J/Cmol), which is the amount of heat required to heat up 1 mol of a substance by one degree).


q=C(ΔT) or q=mc(ΔT) or q=nC(ΔT)\boxed{q=C\left(\Delta T\right)}\ or\ \boxed{ q=mc\left(\Delta T\right)} \ or\ \boxed{q=nC\left(\Delta T\right)}

q is heat (in J)
C is the heat capacity (J/oK or J/oC)
c is the specific heat capacity (J/goK or J/goC)
m=mass (in g)
ΔT=change in temperature (T2-T1) in oK or oC
n=moles

Note: mc=C
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General Introduction to Calorimeters

Calorimeters come in two general varieties:
  • Simple "coffee cup" calorimeters
  • And bomb calorimeters
Left Photo by Community College Consortium for Bioscience Credentials / CC BY, right photo by Lisdavid89 / CC BY

What is a calorimeter?
  • A calorimeter is just a container that is insulated. It holds a liquid that is usually water inside of it and you can have reactions happening inside of it as well.
  • For a calorimeter, the water is the surroundings and the reaction is the system.
  • A thermometer will be able to read the temperature changes of the water.
  • If the thermometer measures an increase in temperature, that means the water went up in temperature.
  • If the water went up in temperature then that means it gained all of the heat from the system (or the reaction taking place inside the calorimeter). It didn't gain the heat from anywhere else because the calorimeter is insulated!
  • This means the system/reaction lost heat and is endothermic/exothermic:
    exothermic.

Wize Concept
In general:
qwater=-qrxn where q=mcΔT

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Coffee Cup Calorimeters

Essentially it's an open styrofoam cup with a thermometer.


These are most frequently used to measure the temperature change of an aqueous reaction under constant
pressure
conditions.

ΔH= ΔE + PΔV
ΔH= q + w -w
ΔH=qp (under constant pressure conditions)


Wize Concept
In general:
qsystem(sample) = -qsurroundings(H2O)
using q=mcΔT:
(mass of sample)(c of sample)(ΔT of sample) = -(mass of H2O)(c of H2O)(ΔT of water)

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Bomb Calorimeters

Typically these consist of a sealed reaction vessel loaded with a combustible material and filled with oxygen placed in a surrounding water bath.

A combustion reaction is performed under constant
volume
conditions and the resulting change in temperature of the water bath is measured.
ΔE= qv + w
ΔE=qv -PΔV
(ΔV=0) so w=0
ΔE=qv

For bomb calorimeters, we will see two types of problems:
1) Sometimes we evaluate the heat capacity for the entire calorimeter (the q of the calorimeter represents bomb and the water bath)

2) Sometimes we evaluate the heat capacity of the bomb and water separately (qtotal=qbomb + qwater bath)

Wize Tip
Read the question carefully to figure out which one of these types of problems we are looking at!


Wize Concept
In general:
qrxn = -qcalorimeter
Can usually use q=CΔT for the bomb calorimeter

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Example: Copper Weight Calorimetry

A copper weight (52.3 g52.3\ g) is heated by a flame from room temperature (25° C25\degree\ C) to a temperature of 189° C189\degree\ C. The specific heat capacity of copper is 0.385 J g−1K−10.385\ J\ g^{-1} K^{-1}

a) What is ΔH\Delta H for this process?

ΔH=qp=mCpΔT=(52.3 g)(0.385 J g−1𝐾−1)(189−25)=3302 J\Delta H = q_p = mC_p\Delta T = (52.3\ g)(0.385\ J\ g^{-1}𝐾^{-1})(189 − 25) = 3302\ J

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b) The block is then allowed to cool in some water until the temperature of the block is 75° C75\degree\ C, what is ΔH\Delta H for this process?

ΔH=qp=mCpΔT=(52.3 g)(0.385 J g−1𝐾−1)(75−189)=−2295 J\Delta H = q_p = mC_p\Delta T = (52.3\ g)(0.385\ J\ g^{-1}𝐾^{-1})(75 − 189) = -2295\ J

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c) What is the ΔH\Delta H from part b) with respect to the water ?

The heat was transferred from the copper to the water so
qwater=−qCu=+2295 Jq_{water}=-q_{Cu}=+2295\ J
Remember that the amount of heat the system (our copper-the thing that is reacting) gains or loses will be the opposite for the surroundings.
In b) we found the system (copper) lose 2295J.
This means the surroundings must gain 2295J.
Since a calorimeter is insulated, whatever is lost by the system is gained by the surroundings and when the surroundings loses energy, it loses that energy to the system!
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Example: Calorimetry

A block of 20 g20\ g lead cube (cp=0.128 J g−1K−1c_p=0.128\ J\ g^{-1}K^{-1}) is heated to 145° C145\degree\ C. The cube is then submerged into a Styrofoam cup containing 500 g500\ g of water at room temperature (cp=4.18 J g−1K−1c_p=4.18\ J\ g^{-1}K^{-1}). After the water and the lead cube come to thermal equilibrium, what is the final temperature of the water?

qp,lead=−qp,watermCpΔT=−mcpΔT(20cg)(0.128 J g−1𝐾−1)(T2−418 K)=−(500 g)(4.18 J g−1K−1)(T2−298 K)(2.56 J K−1)T2−1070.08 J=(−2090 J K−1)T2+622820 J(2.56 J K−1)T2+(2090 J K−1)T2=622820 J+1070.08 JT2(2.56J K−1+2090 J K−1)=622820 J+1070.08 JT2=1070.88 J+622820 J(2.56 J K−1+2090 𝐽 K−1)=298.15 K\begin{array}{c} q_{p,lead}=-q_{p,water}\\[10pt] mC_p\Delta T=-mc_p\Delta T\\[10pt] (20cg)(0.128\ J\ g^{-1}𝐾^{-1})(T_2 - 418\ K) = -(500\ g)(4.18\ J\ g^{-1}K^{-1})(T_2 - 298\ K)\\[10pt] (2.56\ J\ K^{-1})T_2 - 1070.08\ J = (-2090\ J\ K^{-1} )T_2 + 622820\ J\\[10pt] (2.56\ J\ K^{-1})T_2+(2090\ J\ K^{-1} )T_2= 622820\ J + 1070.08\ J \\ \\ T_2(2.56 J\ K^{-1}+2090\ J\ K^{-1})=622820\ J + 1070.08\ J\\ \\ T_2=\dfrac{1070.88\ J + 622820\ J}{(2.56\ J\ K^{-1} + 2090\ 𝐽\ K^{-1})}=298.15\ K \end{array}
Wow! Water is a pretty good insulator
The temperature of a 12.58 g sample of calcium carbonate (CaCO3(s)) increases from 23.6oC to 38.2oC. If the specific heat capacity of calcium carbonate is 0.82 J/g K. How many joules of heat are absorbed?
The ΔHsoln for the process when solid sodium hydroxide (NaOH) dissolves in water is 44 kJ/mol. When a 10.0 g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0oC to _______oC. Assume that the solution has the same specific heat as liquid water, i.e. 4.18 J/gK.

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Example: Bomb Calorimetry

When 2g of rocket fuel, N2H4 is burned inside of a bomb-calorimeter, the temperature of the water rises from 22oC to 29oC. If there is 0.9kg of water and the heat capacity for the bomb is 822J/oC, what is the heat of the combustion for one mole of N2H4 in the bomb calorimeter?

a) 5000 J
b) -51 000 J
c) -513 kJ
d) -573 kJ

Let's first write out all the information we have:


Rocket fuel (N2H4):
m=2g

Bomb Calorimeter:
T1=22oC of water bath
T2=29oC of water bath
ΔT=29-22=7oC
0.9 kg water =900g water in water bath
C(bomb)=822J/oC

This question is asking us to find how much heat (q) is involved when we combust 1 mole of the rocket fuel (N2H4).
We currently are given the situation with 2g of rocket fuel.

We can find how much heat is involved when we have 2g of rocket fuel first. Then we can figure out how much heat would be involved for a mole of the rocket fuel!

For the bomb calorimeter, notice we are only given C(bomb) not C(calorimeter), which would include the bomb and the water bath.
Therefore, the qtotal for the calorimeter = qbomb + qwater bath in this case

qbomb=C(bomb)ΔT
qbomb=822J/oC(7oC)
qbomb= 5754 J

qwater bath=mcΔT
qwater bath=900g(4.18J/goC)(7oC)
qwater bath=26334 J

qtotal for the calorimeter = qbomb + qwater bath
qtotal for the calorimeter = 5754 J + 26334 J
qtotal for the calorimeter = 32 088 J

Let's double check the sign (+ or -) of our q calorimeter. We are told that the water bath temperature increases. This means the water bath (and calorimeter in general) gained heat and so the q value should be +!
qcalorimeter = +32 088J

So in this reaction, the qcombustion reaction= -32 088J
Recall: the reaction is the system and the calorimeter is the surroundings. qsys = -qsurr

Now we know that 32 088J was released when we did this reaction with 2g of our rocket fuel (N2H4).

We need to figure out how many J would be released when we instead use 1 mol of rocket fuel:


1) Figure out how many moles we have when we used 2g

n=m/M M=N2H4= 32g/mol
n=2g/32g/mol
n=0.0625 moles

2) When we had 0.0625 moles of rocket fuel 32 088 J of heat was released

When we have 1 mole of rocket fuel _________ J of heat would be released


1/0.0625 is 16. To double check, check that 0.0625 x16 =1.. it does!
This means we can do 32 088J x16 to get our answer: 513408J would be released if we had 1 mol of rocket fuel!

=513kJ

Practice: Bomb Calorimetry

A fuel compound was burned in oxygen in a bomb calorimeter with a combined heat capacity of 56.5 k]/K (Bomb assembly and water). The temperature of the surrounding water bath increased from 23.5 to 34.6 °C as a result. What was the value of qsys for this process?

Extra Practice
Calorimetry: Heat of combustion
When 2g of rocket fuel, N2H4 is burned inside of a bomb-calorimeter, the temperature of the water rises from 22oC to 29oC. If there is 0.9kg of water and the heat capacity for the bomb is 822J/oC, what is the heat of the combustion for one mole of N2H4 in the bomb calorimeter?
Calorimetry: Heat of combustion
When 2g of rocket fuel, N2H4 is burned inside of a bomb-calorimeter, the temperature of the water rises from 22oC to 29oC. If there is 0.9kg of water and the heat capacity for the bomb is 822J/oC, what is the heat of the combustion for one mole of N2H4 in the bomb calorimeter?
Heat Capacity (bomb calorimeter)
1 gram of fried chicken is placed in a bomb calorimeter, which has a heat capacity of 10 kJ/oC. Upon combustion of the fried chicken, the temperature of the calorimeter rises by 5oC. Approximately how many nutritional calories are in 1 gram of this fried chicken if 1 nutritional calorie = 1000 calories and 1 calorie = 4.184 J ? Enter answer with 2 significant digits
Calorimetry
Equal masses of copper and iron are both heated to a temperature of 100°C. Each piece of hot metal is then placed into its own beaker containing 1 L of water. If both beakers start with the same initial temperature, which beaker of water will have the higher final temperature?
Additional information: Cu (specific heat = 0.377 J/g°C); Fe (specific heat = 0.450 J/g°C)
Calorimetry
The temperature of a 12.58 g sample of calcium carbonate (CaCO3 (s)) increases from 23.6oC to 38.2oC. If the specific heat of calcium carbonate is 0.82 J/g K, how many joules of heat are absorbed?
Calorimetry: Heat requirement
How much heat is required to convert a 15.5 g ice cube at -5.00oC to liquid water at 0oC? (The specific heat of ice is 2.09 J/g K, the specific heat of water is 4.18 J/g K, and the ΔHfus for water=334 J/g).
Calorimetry: Specific Heat Capacity
Addition of 3.10 kJ of heat is required to raise the temperature of 150 g of a liquid hydrocarbon from 23.0oC to 44.0oC. What is the specific heat capacity of this hydrocarbon?
A bomb calorimeter assembly has a heat capacity of 650 J/°C. Inside the bomb calorimeter 2.5g of a fuel compound is burned in excess oxygen and the temperature of the bomb calorimeter increases by 48 °C. What is the value for qrxn amount of heat released per gram of compound burned?
Please report your answer to three significant figures in kJ/g. Do NOT include units in the answer field
Water can absorb infrared radiation at a wavelength of1.06×104 m{\small 1.06\times10^4 \text{ m}}. If a 180 g sample of water is irradiated at this wavelength, calculate the number of photons required to raise the temperature by 25.0°C(CH2O=4.184Jg°C){\small \degree \text{C} \left(C_{H_2O} =4.184\frac{J}{g\degree C} \right)}
Calorimetry: Heat of combustion
When 2g of rocket fuel, N2H4 is burned inside of a bomb-calorimeter, the temperature of the water rises from 22oC to 29oC. If there is 0.9kg of water and the heat capacity for the bomb is 822J/oC, what is the heat of the combustion for one mole of N2H4 in the bomb calorimeter?
Enthalpy: Heat of reaction
2H2(g) + O2(g) --> 2H2O(l) ΔH=-572kJ/mol
1. How much heat is produced when 72g of water gas is produced?
2. Is the reaction endothermic or exothermic?
When 1.00 kg of lead (specific heat = 0.13 J / g °C) at 125 °C is added to a quantity of water at 18.0 °C, the final temperature of the lead-water mixture is 32.0 °C. What is the volume of water present? Note: the density of H2O(l) is 1.00 g/mL and the specific heat capacity is 4.18 J / g oC.
Calorimetry: Specific Heat
A piece of metal weighing 58g was heated to 100oC and then added to 100mL of water. The water's temperature was initially 22oC and after the metal was added the temperature was 26oC. What is the specific heat of the metal?