Calorimetry

LOCOE

(review)

The Law of Conservation of Energy

Energy cannot be created or destroyed, only transferred.

* this is essentially what we do in calorimetry

If we place the hot block of iron into water, the heat from the iron will be released and absorbed by the water. This is a transfer of energy.

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We represent this using the following equations:

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q_sys + q_sur = 0 - q_sys = q_sur q_sys = - q_sur

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Calorimetry

New Concept!

What if we want to know how much heat is moving from the block of iron into the water?

We use calorimetry!

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Calorimetry is the measurement of heat (q) flowing into or out of a system. It uses a tool called a calorimeter which is an insulated container that is used to measure heat transfer.

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Calorimeters

There are two main types of calorimeters:

A bomb calorimeter and a coffee cup calorimeter

Both calorimeters function the same way and are built with the same basic structure: an insulated container that can be filled with liquid, a thermometer, and a stirrer.

The container must be insulated in order to limit heat transfer between the liquid in the chamber and the air in the surroundings.

Calorimetry Videos

Another review video by Hank Green!

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Start at 3:06

End at 9:10�

Calorimetry In-class Practice

A metal is heated and placed in a calorimeter containing 14.0g of water at 24.7°C. The water (the surroundings) reaches a maximum temperature of 40.2°C. Calculate the heat absorbed by the surroundings. C_water = 4.184 J/g°C

Practice #1

Use q = m·C·ΔT

To calculate the q of the surroundings (water)

q_water = 14.0g·4.184J/g^oC·15.5^oC

Calculate ΔT

ΔT = T_f - T_i

ΔT = 40.2^oC - 24.7^oC

ΔT = 15.5^oC

q_water = 907.928 J

A metal is heated and placed in a calorimeter containing 14.0g of water at 24.7°C. The water (the surroundings) reaches a maximum temperature of 40.2°C. Calculate the heat absorbed by the surroundings. C_water = 4.184 J/g°C

A metal is heated and placed in a calorimeter containing 14.0g of water at 24.7°C. The water (the surroundings) reaches a maximum temperature of 40.2°C. Calculate the heat absorbed by the surroundings. C_water = 4.184 J/g°C

Practice #1

Remember, the water is absorbing 907.928 J from the metal. So, how much heat is the metal releasing?

q_water = 907.928 J

A metal is heated and placed in a calorimeter containing 14.0g of water at 24.7°C. The water (the surroundings) reaches a maximum temperature of 40.2°C. Calculate the heat absorbed by the surroundings. C_water = 4.184 J/g°C

So, - 907.928 J is being released from the metal.

Use LOCOE: q_sys + q _sur= 0

2. 4.5 grams of a metal is heated to 82.5 °C and placed in a calorimeter containing 14.0 g of water at 24.7 °C. The water and metal reach a maximum temperature of 40.2 °C.

Calculate the specific heat of the metal (C_metal). C_water = 4.184 J/g^oC

← These are the same value, always!

4.5 g

???????

82.5 ^oC

40.2 ^oC

40.2 ^oC

24.7 ^oC

4.184 J/g^oC

14.0 g

15.5 ^oC

-42.3 ^oC

907.928 J

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-907.928 J

metal

water

1. Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 40.2^oC - 24.7^oC 𝛥T _system = 40.2^oC - 82.5^oC

𝛥T _surroundings = 15.5 ^oC 𝛥T _system = - 42.3^oC

q _sur= 14.0g (4.184J/g^oC)(15.5^oC) q_sur = 907.928 J

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0 q_sys + 907.928 J = 0 q_sys = - 907.928 J

• Plug values for the system or surroundings into q=mC𝛥T and solve for C.

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- 907.928 J = 4.5g (x)(- 42.3^oC)

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C_sys = 4.7697819 → 4.8 J/g^oC

4.8 J/g^oC

Calorimetry Practice

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Walkthrough Slides

Calorimetry

Practice

(question #3)

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• A 3.00x10² g piece of iron was heated to 87.5°C and placed in a calorimeter that contained 123 g of water at 10.0 °C. The final temperature of the iron and water was 60.5^o C. Calculate the specific heat of iron. The specific heat for water is 4.184 J/g^oC.

← These are the same value, always!

300. g

???????

87.5 ^oC

60.5 ^oC

60.5 ^oC

10.0 ^oC

4.184 J/g^oC

123 g

50.5 ^oC

-27.0 ^oC

25,988.916 J

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-25,988.916 J

iron

water

• Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 60.5^oC - 10.0^oC 𝛥T _system = 60.5^oC - 87.5^oC

𝛥T _surroundings = 50.5 ^oC 𝛥T _system = -27^oC

q _sur= 123g (4.184J/g^oC)(50.5^oC) q_sur = 25,988.916 J

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0 q_sys + 25,988.916 J = 0 q_sys = -25,988.916 J

• Plug values for the system or surroundings into q=mC𝛥T and solve for C.

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-25,988.916 J = 300.g (x)(-27^oC)

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C_sys = 3.2085 → 3.21 J/g^oC

3.21 J/g^oC

VIDEO WALKTHROUGH

Calorimetry

Practice

(question #4)

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4. A piece of metal with a mass of 59.04 g was heated to 100.0 °C and then put into a coffee cup calorimeter containing 100.0 g of water (initially at 23.7 °C). The metal and water were allowed to come to thermal equilibrium, meaning the final temperature for both substances (system and surroundings) was 27.8 °C. What is the specific heat of the unknown metal? The specific heat of water is 4.184 J/g^oC.

← These are the same value, always!

59.04 g

???????

100.0 ^oC

27.8 ^oC

27.8 ^oC

23.7 ^oC

4.184 J/g^oC

100.0 g

4.1 ^oC

-72.2 ^oC

1715.44 J

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-1715.44 J

metal

water

• Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 27.8^oC - 23.7^oC

𝛥T _surroundings = 4.1 ^oC

q _sur= 100.0g (4.184J/g^oC)(4.1^oC)

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0

• Plug values for the system or surroundings into q=mC𝛥T and solve for C.

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-1715.44 J = 59.04g (x)(-72.2^oC)

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𝛥T _system = 27.8^oC - 100.0^oC

𝛥T _system = -72.2^oC

-1715.44 J = -4262.688 (x)

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0.402 J/g^oC

q_sur = 1715.44 J

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q_sys = -1715.44 J

q_sys + 1715.44 J = 0

C_sys = 0.4024315174 → 0.402 J/g^oC

VIDEO WALKTHROUGH

Calorimetry

Practice

(question #5)

5. A piece of iron (C_iron = 0.460 J/g×°C) with unknown mass was heated to 87.5° C and placed in a calorimeter that contained 123 g of water (C_water = 4.184 J/g×°C) at 10.0 ° C. The final temperature of both the iron and water was 60.5° C. Calculate the mass of iron used.

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← These are the same value, always!

???????

0.460 J/g^oC

87.5 ^oC

60.5 ^oC

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60.5 ^oC

10.0 ^oC

4.184 J/g^oC

123 g

50.5 ^oC

-27.0 ^oC

25,988.913 J

-25,988.913 J

Iron

water

• Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 60.5^oC - 10.0^oC

𝛥T _surroundings = 50.5 ^oC

q _sur= 123g (4.184J/g^oC)(50.5^oC)

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0

• Plug values for the system or surroundings into q=mC𝛥T and solve for m.

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-25,988.913 J = x (0.46 J/g^oC)(-27.0^oC)

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𝛥T _system = 60.5^oC - 87.5^oC

𝛥T _system = -27.0 ^oC

-25,988.913 J = -12.42(x)

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2090 g

q_sur = 25,988.913 J

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q_sys = -25,988.913 J

q_sys + (25,988.913 J) = 0

m_sys = 2092.5050 → 2090 g

VIDEO WALKTHROUGH

Calorimetry

Practice

(CHALLENGE question #6)

6. CHALLENGE: A piece of copper (C_copper = 0.092 cal/g×°C) of mass 19.0 g was heated to 87.4° C and placed in a calorimeter that contained 55.5 g of water at 18.3° C. The final temperature of both the water and the copper was 20.4° C. Calculate the specific heat of water from this process.

← These are the same value, always!

19.0 g

0.092 cal/g^oC

87.4 ^oC

20.4 ^oC

20.4 ^oC

18.3 ^oC

???? cal/g^oC

55.5 g

2.1 ^oC

-67.0 ^oC

117.116 cal

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-117.116 cal

copper

water

• Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 20.4^oC - 18.3^oC

𝛥T _surroundings = 2.1 ^oC

q _sys = 19.0g (0.092cal/g^oC)(-67.0^oC)

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0

• Plug values for the system or surroundings into q=mC𝛥T and solve for C.

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117.116 cal = 55.5g (x)(2.1^oC)

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𝛥T _system = 20.4^oC - 87.4^oC

𝛥T _system = -67.0^oC

117.116 cal = 116.55 x

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1.00 cal/g^oC

q_sur = -117.116 cal

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q_sur = 117.116 cal

- 117.116 cal + q _sur = 0

C_sys = 1.00485 → 1.00 cal/g^oC

VIDEO WALKTHROUGH

Extra Calorimetry

Practice

Problems

7. Santa Claus is very fussy about his bathwater temperature. It has to be exactly 38.0^° C. One day Mrs. Claus notices that it is 42.0^° C. To cool the 1.00×10³ g of bathwater to 38.0^° C she plans on adding an aluminum ducky to the bathwater. If the aluminum ducky was originally at freezer temperature (-24.0^° C) what mass should the aluminum ducky be so that both the water and ducky’s final temperature is 38.0^° C?

(C_water = 4.184 J/g×°C, C_Al = 0.900 J/g×°C).

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← These are the same value, always!

???????

0.900 J/g^oC

-24.0 ^oC

38.0 ^oC

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38.0 ^oC

42.0 ^oC

4.184 J/g^oC

1000 g

-4.0 ^oC

62.0 ^oC

-16,736 J

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16,736 J

Al Ducky

bathwater

• Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 38.0^oC - 42.0^oC

𝛥T _surroundings = -4.0 ^oC

q _sur= 1000g (4.184J/g^oC)(-4.0^oC)

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0

• Plug values for the system or surroundings into q=mC𝛥T and solve for m.

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16,736 J = x (0.900 J/g^oC)(62.0^oC)

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𝛥T _system = 38.0^oC - (-24.0^oC)

𝛥T _system = 62.0^oC

16,736 J = 55.8(x)

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300. g

q_sur = -16,736 J

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q_sys = 16,736 J

q_sys + (-16,736 J) = 0

m_sys = 299.9283 → 300. g

VIDEO WALKTHROUGH

8. An unknown solid with a mass of 14.4 g was heated to 94.5° C and placed in a calorimeter that contained 60.1 g of water (C_water = 4.184 J/g×°C) at 20.5° C. The final temperature of both the water and the unknown solid was 30.5° C. Calculate the specific heat of the unknown solid (C_solid) from this process.

← These are the same value, always!

14.4 g

???????

94.5 ^oC

30.5 ^oC

30.5 ^oC

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20.5 ^oC

4.184 J/g^oC

60.1 g

10.0 ^oC

-64.0 ^oC

2514.584 J

-2514.584 J

Unknown solid

water

• Determine your system and surroundings.

• Fill out as much of the chart as possible using the numbers from the word problem. Use the units as a way of identifying which number belongs to which variable.

• Plug values for the system or surroundings into q=mC𝛥T and solve for q.

• Calculate 𝛥T for both the system and the surroundings using 𝛥T = T_f - T_i

𝛥T _surroundings= 30.5^oC - 20.5^oC

𝛥T _surroundings = 10.0 ^oC

q _sur= 60.1g (4.184J/g^oC)(10.0^oC)

• Use LOCOE to find the amount of heat the system/surroundings absorbed or released.

q_sys + q _sur= 0

• Plug values for the system or surroundings into q=mC𝛥T and solve for C.

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-2514.584 J = 14.4g (x)(-64.0^oC)

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𝛥T _system = 30.5^oC - 94.5^oC

𝛥T _system = -64.0^oC

-2514.584 J = -921.6 (x)

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2.73 J/g^oC

q_sur = 2514.584 J

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q_sys = -2514.584 J

q_sys + 2514.584 J = 0

C_sys = 2.7284982 → 2.73 J/g^oC

VIDEO WALKTHROUGH