Calorimetry lesson and practice problems

Calorimetry

By Tania Murphy

calorimetry is the process of measuring the amount of heat released or the amount of heat absorbed in a chemical reaction or physical proces

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​ What is calorimetry?

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A key component to calorimetry is understanding the energy is neither created nor destroyed but is conserved in all energy transformations. Therefore, the energy lost by the system will be ​

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the energy gained by the ​

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and vice versa. If water absorbed 200 J of energy from a hot metal then the hot metal must have lost ​

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of energy. If the water released 30 J of energy to the metal then the metal gained ​

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of energy. This is due to the ​

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equal to

surroundings

200 J

30 J

Law of Conservation of Energy

less than

greater than

Law of Conservation of Mass

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Another key component, especially when using calorimetry to determine the specific heat of a metal is to understand thermal equilibrium. When a hot object and a cold object are combined, ​

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will flow from the ​

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to the ​

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until both reach the ​

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. We call this ​

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heat

hot object

cold object

same temperature

thermal equilibrium.

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Thus, if a hot piece of metal is placed in a sample of room temperature water, the temperature of the water will ​

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and the temperature of the metal will ​

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. The final temperature of the water will be ​

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the final temperature of the metal. If a metal at 90 degree Celsius is placed in a sample of water and the temperature of the water rises from 25 celsius to 35 degrees Celsius, the final temperature of the metal will be ​

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cool down

heat up

equal to

less than 

greater than

35 degree Celsius

greater than 35 degree Celsius 

less than 35 degree Celsius

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Another thing to consider is water has a density of 1.00g/mL. Therefore, 100.0 mL of water has a mass of ​

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, 55.0 mL of water has a mass of ​

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etc.

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100.0 g

55.0g

Math Response

The next several slides will deal with the following scenario. Write your work down on a piece of paper as you go. We will take a stepwise approach to solving the problem. A 58.8 g piece of hot alloy (metal) is placed in a calorimeter that contains 125 g of cold water. The temperature of the alloy decreases by 106.1 C, while the temperature of the water increases by 10.5 C. The specific heat of water is 4.184 J/gC. What is the specific heat of the alloy? Step 1: Calculate the heat absorbed by the water using

q=m c deltaT

enter with one digit behind the decimal and do not put a unit

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Correct Answer: 5491.5

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The water gained the heat from ​

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and the heat gained by the water is ​

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the heat lost by the metal. This provides you with the information you need to complete step 2 which is finding the specific heat of the metal. Because you now know the heat lost by the metal, the mass of the metal and the temperature change of the metal.

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the metal

the beaker

the thermometer

equal to

Math Response

Same question continued. Use the heat gained by the metal, the mass of the metal and the change in temperature of the metal to calculate the specific heat of the metal. Type in your answer with three digits after the decimal and no units.

Type answer here

Correct Answer: 0.880

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A 20.50 g sample of a metal is placed in a beaker of water that is boiling at 100 degrees Celsius. The hot metal is then quickly removed from the boiling water and placed in 105.0mL of water at a temperature of 25.6 degrees Celsius. The final temperature of the water is 28.7 degrees Celsius.



The mass of the sample of water is​

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The initial temperature of the metal is ​

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and the final temperature of the metal is ​

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. Write all of your "facts" down and the next few slides will walk you through determining the specific heat of the metal

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105.0 g

28.7 degrees Celsius

100 degrees Celsius

Math Response

A 20.50 g sample of a metal is placed in a beaker of water that is boiling at 100 degrees Celsius. The hot metal is then quickly removed from the boiling water and placed in 105.0mL of water at a temperature of 25.6 degrees Celsius. The final temperature of the water is 28.7 degrees Celsius. We will take a step wise approach to determining the specific heat of the metal. (the specific heat of water is 4.19=84 J/gC)

Step 1: Calculate the heat gained by the water and enter here (round to the tenth place

Type answer here

Correct Answer: 1361.9

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Math Response

Use your previous answer to calculate the specific heat of the metal and type in your answer (round to the thousandth place)

Type answer here

Correct Answer: 0.932

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A piece of metal with a mass of 3.99 g and an initial temperature of 100.0 degrees Celsius is placed into 59.0 mL of water at a temperature of 21.0 degrees Celsius causing the temperature of the water to rise 2.5 degrees Celsius. The mass of the water is​

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and the final temperature of the water and metal is 23.5 degrees Celsius. The change in temp for the water is ​

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and the change in temp of the metal is ​

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. Write down all of your facts and use them to answer the question on the next slide.

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59.0g

2.5 degrees Celsius

76.5 degrees Celsius

Math Response

A piece of metal with a mass of 3.99 g and an initial temperature of 100.0 degrees Celsius is placed into 59.0 mL of water at a temperature of 21.0 degrees Celsius causing the temperature of the water to rise 2.5 degrees Celsius. What is the specific heat of the metal? (round to two digits after the decimal)

Type answer here

Correct Answer: 2.02

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Math Response

You will complete this problem on your own. A 75.0 g metal is in boiling water at 100.0 degrees Celsius. The metal is quickly removed and placed into 150 mL of water at 25 degree Celsius causing the temperature of the water to rise to 26.8 degrees Celsius. What is the specific heat of the metal?

(round to the thousandth place)

Type answer here

Correct Answer: 0.206

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