Practice Sheet 17 Answers PDF

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Practice Sheet 17 Answers...

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Practice Sheet 17 and Answers Please use these answers (at the end) to make sure you understand and know how to do the problems and to study for the exam. Practice Problem 17D.1. (a) What is the specific heat of liquid water? (b) What is the molar heat capacity of liquid water? (c) What is the heat capacity of 185 g of liquid water? (d) How many kJ of heat are needed to raise the temperature of 10.00 kg of liquid water from 24.6 °C to 46.2 °C? 17D.2. When a sample of gold is cooled from 32.5 °C to 20.0 °C, 9.46 J of heat are liberated. What is the mass of the sample if the specific heat of gold is 0.129 J/g·°C?

Parallel Example Define the following terms a) specific heat b) heat capacity c) molar heat capacity.

Key to Parallel Example a) The amount of energy required to raise one gram of a substance by one degree Celsius/Kelvin. b) The amount of energy required to raise a specified amount (usually by mass) of a substance by one degree Celsius/Kelvin. c) The amount of energy required to raise one mole of a substance by one degree Celsius/Kelvin.

Pure aluminum metal (3.79 g) is cooled by water from 103.2 °C to 34.8 °C. The heat capacity of aluminum is 0.890 J/g⋅°C. If the water started out at 32.7 °C, how much water was used to cool the aluminum?

The heat lost by the aluminum is equal to the heat gained by the water. The final temperatures of the water and the aluminum are the same. Again, we use the equation q = mc∆T, but this time to find the mass of the water. Using the answer from the 16 BR example problem 1 (231 J) we can find the mass of the water used to cool the aluminum.

mw =

qw 231 J = = 26.3 g H2O J c sw ΔT 4.184 g ⋅°C (34.8°C − 32.7°C)

(

)

17D.3. A sample of iron metal is cooled by 1000. g of water. The temperature of the water is raised from 23.0 °C to 30.1 °C. Before contact with the water, the temperature of the iron was 40.9 °C. What was the mass of the iron sample? The specific heat capacity of iron is 0.449 J/g⋅°C. (Assume the final temperature of the iron and water are the same.) 17D.4. Pondering the concept of specific heat while walking home from chemistry class, you find an old penny on the ground and begin to wonder what the specific heat of copper is. You rush home and put 5.00 g of water at 18.00 °C in a styrofoam cup. You heat the 2.45 g penny to 50.00 °C and drop it in the water. The final temperature of the metal and water is 19.38 °C. Assuming the penny is pure copper (which modern pennies are not), what is the specific heat of copper? The specific heat of water is 4.18 J/g· °C.

A 78.0 g aluminum bar at a temperature of 25.0 °C is placed in 225 mL (225 g) of hot water at a temperature of 95.0 °C. Assuming no heat is lost to the surroundings, what will be the final temperature of the water and aluminum bar? The specific heat of aluminum and water are 0.88 J/g·°C and 4.18 J/g·°C, respectively.

Link to video of this problem.

In this process, the heat gained by the aluminum equals the heat lost by the water. We know that the amount of heat transferred by the substance can be determined by the equation q = mcs∆T, where m is the mass of the substance, cs is the specific heat, and ∆T is the temperature change of the substance. Thus, we can set the following equivalency. mAlcs(Al)∆TAl = −mwcs(w)∆Tw The change in temperature of the water and aluminum (∆T) is equal to the final temperature minus the initial temperature, so we can rewrite the equation. mAlcs(Al)(Tfinal− TiA l ) = −mwcs(w)(Tfinal − T i w ) Note that the final temperature of the aluminum and the water is the same. Substituting and solving for the final temperature gives us the answer. We also will use temperature in °C instead of K because the J units of specific heat are g C . °

(78.0

g)(0.88 J g ⋅ C

)(Tf − 25.0 °C) = − (225 g)(4.18

(68.6

J C

)(Tf − 25.0 °C) = −(941

68.6

J C

Tf − 1.72×103 J = −941

68.6

Tf = 90.2 °C

J C

J C J C

J g ⋅ C

)(Tf − 95.0 °C)

)(Tf − 95.0 °C)

Tf + 8.94×104 J

Tf + 941

J C

Tf = 8.94×104 J + 1.72×103 J

1010

J C

Tf = 9.11×104 J

Practice Sheet 17 Answers 17D.1 a) 4.184 J/g·K b) (4.184 J/g·K) × (18.02 g/mol) = 75.40 J/mol·K c) (185 g) × ( 4.184 J/g·K) = 774 J/K  1000 g  4.184 J  d) q water = mc s ∆T = 10.0 kg    (46.2 − 24.6) K = 904,000 J =  kg  g ⋅ K  904 kJ 17D.2 We can use the equation qgold = mgoldcs(gold)∆Tgold. Solving for the mass of gold and substituting for the other variables gives us the answer. Remember, qgold is negative because heat was given off by the gold. mgold =

q gold c s(gold) ∆Tgold

=

(0.129

J g ⋅ C

− 9.46 J = 5.87 g )(20.0  C − 32.5  C)

17D.3 First we need to solve for qwater which is equal to qFe.  4.184 J  q water = mc s ∆T = 1000 g   (30.1 − 23.0 ) K = 29, 706.4 J  g ⋅K  So, the mass of the iron is given by: q Fe − 29, 706.4 J m Fe = = = 6126 g = 6.13 kg c s(Fe) ∆TFe (0.449 J / g ⋅ K )(30.1− 40.9) K 17D.4 In this process, the heat gained by the water equals the heat lost by the penny. (In equation form: qwater = −qpenny.) We know that the amount of heat transferred by a substance can be determined by the equation q = mcs∆T. Thus, we can set the following equivalency. mwatercs(water)∆Twater = − mpennycs(penny)∆Tpenny Substituting and solving for cs(penny) gives us the answer.

cs(penny) =

mwater c s(water) ∆Twater − m penny ∆ Tpenny

(5.00 g)(4.18 =

J g⋅  C

)(19.38  C − 18.00  C)

− (2.45 g)(19.38  C − 50.00  C)

= 0.384 J / g ⋅ °C...