How Do I Solve These Chem Problems? (Physical Thermochemistry) ?

Question

My chemistry teacher was not good during distance learning at the end of last year so I have no idea what to do for these problems in my AP chem summer work.

1. A 15.3g piece of metal with an initial temp of 100.9°C is added to 90.3g of water with an initial temp of 20.9°C. The final temp is 27.0°C. Calculate the specific heat of the metal.

2. A 10.4g piece of metal with an initial temperature of 100°C is added to 33.3g of water with an initial temp of 22.0°C. The specific heat of the metal is 0.593J/g°C. Calculate the final temp. (Cp for water is 4.18 J/g°C)

3. What is the enthalpy of solution in kJ/mole if 5.05g of NaOH is dissolved into 124g of water? The temp of the water changes from 23.1°C to 26.4°C. Is this exothermic or endothermic? Explain.

4. When ice at 0°C melts to liquid at 0°C, it absorbs 0.334 kJ of heat per gram. Suppose the heat needed to melt 38.0 g of ice is absorbed from the water contained in a glass. If this water has a mass of 0.210kg and a temperature of 21.0°C, what is the final temp of the water? (Note that you will also have 38.0g of water at 0°C from the ice.)

Roger the Mole · Accepted Answer

1. First, calculate the amount of heat gained by the water:
(4.184 J/g·°C) x (90.3 g) x (27.0 - 20.9)°C = 2304.67 J gained by the water

The same amount of heat was lost by the metal, so:
(2304.67 J) / (15.3 g) / (100.9 - 27.0)°C = 2.04 J/g·°C (the requested specific heat)

(Although it is difficult to imagine a metal with a specific heat that high which does not react with water.)

2.
Let T be the final temperature (in °C) to be found.
(0.593 J/g°C) x (10.4 g) x (100 - T)°C = (616.72 - 6.1672 T) J lost by the metal

(4.184 J/g·°C) x (33.3 g) x (T - 22.0)°C = (139.327 T - 3065.2) J gained by the water

Set to two expressions for the number of joules lost or gained equal to each other:
616.72 - 6.1672 T = 139.327 T - 3065.2
Solve for T algebraically:
T = 25.3°C

3.
Supposing the specific heat of the solution is the same as for water:
(4.184 J/g·°C) x (5.05 g + 124 g) x (26.4 - 23.1)°C = 1781.82 J produced by the dissolution

And since the temperature rose, the dissolution is exothermic.

(1781.82 J) / (5.05 g NaOH / (39.99715 g NaOH/mol)) = 14112 J/mol = -14.1 kJ/mol

(The final minus sign did not come from the calculations.  It is the convention for exothermic processes like this one.)

4.
First, calculate the amount of heat required to melt the ice:
(334 J/g) x (38.0 g) = 12692 J

Let T be the final temperature (in °C) to be found.

(4.184 J/g·°C) x (38.0 g) x (T - 0)°C = (158.992 T) J required to warm the melted ice to T

(4.184 J/g·°C) x (210 g) x (21.0 - T)°C = (18451.44 - 878.64 T) J total lost by the warm water

Set the expressions for heat required or lost equal to each other (units in joules):
12692 + 158.992 T = 18451.44 - 878.64 T
Solve for T algebraically:
T = 5.55°C

ChemTeam · Answer

1) Here's some discussion and sample problems:

https://www.chemteam.info/Thermochem/Determine-Specific-Heat.html

For your problem:

(15.3 g) (73.9 C) (x) = 90.3 g) (6.1 C) (4.184 J/gC)

2) More discussion/examples:

https://www.chemteam.info/Thermochem/MixingMetal&Water.html

For your problem:

(10.4 g) (100 - x) (0.593J/g°C) = (33.3 g) (x - 22) (4.814 J/g°C)

x is the final temp, not the change.

3) I'm going to assume NaOH specific heat is the same as liquid H2O. It's exothermic because the water temp went up.

q = (129.05 g) (3.3 C) (4.184 J/gC) = 1781.81916 J

5.05 g / 40.0 g/mol = 0.12625 mol

1781.81916 J / 0.12625 mol = 14.113 kJ/mol (round off as you see fit)

This is not the accepted value for the dissolving of NaOH. Here's another problem like yours:

https://socratic.org/questions/chemistry-question

Look at the discussion after the problem is solved.

=========

Here's 4:

(38.0 g) (0.334 kJ) = 12.692 kJ

The ice absorbed this energy and became liquid water at zero degrees.

12692 J = (210 g) (x) (4.184 J/g C)

x = 14.445

21 - 14.445 = 6.555 C

6.555 C is the temp of the 210 g of water after it gave up some heat to melt the ice.

You now have this problem, which I will leave you to work on:

38.0 g of water at zero C is mixed with 210 g of water at 6.555 C. What is the final temp?

I have some discussion and examples here:

https://www.chemteam.info/Thermochem/MixingWater.html

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