Calorimetry Lab - Revision history

Bes: /* Example 1 */

2020-04-08T11:26:12Z

Example 1

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Problem 4

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Problem 2

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Example 1

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Problem 4

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Problem 3

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Problem 1

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Example 1

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Units of Heat

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Units of Heat

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 ====Example 1====
 :Q: If 1 mole (~40 g) of NaOH is dissolved in 1000 mL (1 L) of water, what is the change in the temperature of the water?
-:A:  To answer this question we must introduce the concept of "heat capacity." As stated above, 1 J will raise the temperature of 1 gram of water by 0.24 <sup>o</sup>C. Heat capacity is in units of J/(g<sup>o</sup>C), so the above statement says: 1J/(1 g * 0.24 oC) = 4.16 J/(g<sup>o</sup>C); the currently accepted value for the heat capacity of water is 4.184 J/(g<sup>o</sup>C). So using the following equation (e. 6.12 - Chang/Goldsby):
+:A:  To answer this question we must introduce the concept of "heat capacity." As stated above, 1 J will raise the temperature of 1 gram of water by 0.24 <sup>o</sup>C. Heat capacity is in units of J/(g<sup>o</sup>C), so the above statement says: 1J/(1 g * 0.24 <sup>o</sup>C) = 4.16 J/(g<sup>o</sup>C); the currently accepted value for the heat capacity of water is 4.184 J/(g<sup>o</sup>C). So using the following equation (e. 6.12 - Chang/Goldsby):
 ::::q (heat in J) = (mass of water in grams) * (Specific heat of water) * (ΔT in units of <sup>o</sup>C - change in the temperature in <sup>o</sup>C)
 ::::44,510 J = 1000 g * 4.184 J/(g<sup>o</sup>C) * ΔT, solving for ΔT =  10.6 <sup>o</sup>C.

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 :Q: If 12.3 gram of NaOH is dissolved in 500 mL of water, what is the change in the temperature of the water?
-<div align="right"> A: 8.16 <sup>o</sup>C</div>
+<div align="right"> A: 6.54 <sup>o</sup>C</div>
 ===Review===

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 ====Problem 2====
-:Q: If 0.50 mole of NaOH is dissolved in 500 mL of water, how hot will the water get?
+:Q: If 0.50 mole of NaOH is dissolved in 500 mL of water, what is the change in the temperature of the water?
 <div align="right"> A: 10.6 <sup>o</sup>C</div>

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 ====Example 1====
-:Q: If 1 mole (~40 g) of NaOH is dissolved in 1000 mL (1 L) of water, what is the chang ein the temperature of the water?
+:Q: If 1 mole (~40 g) of NaOH is dissolved in 1000 mL (1 L) of water, what is the change in the temperature of the water?
 :A:  To answer this question we must introduce the concept of "heat capacity." As stated above, 1 J will raise the temperature of 1 gram of water by 0.24 <sup>o</sup>C. Heat capacity is in units of J/(g<sup>o</sup>C), so the above statement says: 1J/(1 g * 0.24 oC) = 4.16 J/(g<sup>o</sup>C); the currently accepted value for the heat capacity of water is 4.184 J/(g<sup>o</sup>C). So using the following equation (e. 6.12 - Chang/Goldsby):
 ::::q (heat in J) = (mass of water in grams) * (Specific heat of water) * (ΔT in units of <sup>o</sup>C - change in the temperature in <sup>o</sup>C)

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 Note: in order to do this problem, you need to convert grams to moles first, then repeat as above...
-:Q: If 12.3 gram of NaOH is dissolved in 500 mL of water, how hot will the water get?
+:Q: If 12.3 gram of NaOH is dissolved in 500 mL of water, what is the change in the temperature of the water?
 <div align="right"> A: 8.16 <sup>o</sup>C</div>

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 :::1 mole/44510 J = 0.5 moles/X J --> solving for X = 0.5 moles * 44510 J/ 1 mole = 22255 J. The value 44510 J/mol is called the enthalpy with the symbol ΔH. Look closely at the units, if you multiple the enthalpy by the number of moles you get J.
-:Q: If 0.321 mole of NaOH is dissolved in 500 mL of water, how hot will the water get?
+:Q: If 0.321 mole of NaOH is dissolved in 500 mL of water, what is the change in the temperature of the water?
 <div align="right"> A: 6.83 <sup>o</sup>C</div>

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 ====Problem 1====
-:Q: If 1.00 mole of NaOH is dissolved in 500 mL (500 g since the density - 1g/ml) of water, how hot will the water get?
+:Q: If 1.00 mole of NaOH is dissolved in 500 mL (500 g since the density - 1g/ml) of water, what is the change in the temperature of the water?
 <div align="right"> A: 21.3 <sup>o</sup>C</div>

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 ::NaOH (s) --> NaOH (aq)
-How much heat (in joules) was generated during this reaction? If you do an internet search on the "heat of dissolution" or "heat of solution" [your book has some of these, but not NaOH (s)] you will find a tabulated value of -44.51 kJ/mol, which is referred to as the "enthalpy of solution" or enthalpy of dissolution or "heat of solution" abbreviated (ΔH) Let's look more closely:
+How much heat (in joules) was generated during this reaction? If you do an internet search on the "heat of dissolution" or "heat of solution" [your book has some of these, but not NaOH (s)] you will find a tabulated value of -44.51 kJ/mol, which is referred to as the "enthalpy of solution" or enthalpy of dissolution or "heat of solution" abbreviated (ΔH). Let's look more closely:
 :- this value is negative, meaning that the process is exothermic (heat is given off by NaOH when dissolved in water, '''''the heat goes into the water''''').
 :- the value is is kJ (kilojoules) not J (joules).