rockerchick26 Posts: 93, Reputation: 22 Junior Member #1 Nov 29, 2009, 03:13 PM
Thermodynamics/Calorimetry Practice Problems
I was wondering if someone might be able to check my answers to some thermo questions:

1. A 625 mL sample of water was cooled from 50 C to 10 C. How much heat was lost?

2. How many joules are required to change the temperature of 60g of water from 23.3 C to 38.3 C?

3. Calculate the final temperature when 50 mL of water at 60 C are added to 25 mL of water at 20 C.

4. A piece of metal weighing 5.10 g at a temperature of 48.6 C was placed in a calorimeter containing 20.00 mL of water at 22.1C, and the final equilibrium temperature was found to be 28.2 C. What is the specific heat of the metal?

5. If the specific heat of methanol is 2.51 J/K*g, how many joules are necessary to raise the temperature of 250 g of methanol from 18 C to 33 C?

6. When a 3.25g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 C to 32.0 C. Calculate deltaH (in kJ/mol NaOH) for the solution process:

NaOH (s) ----> Na+(aq) + OH (aq)

Assume it's a perfect calorimeter and that the specific heat of the solution is the same as that of pure water.

:D
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Dec 1, 2009, 06:45 AM

1. If you take the specific heat capacity of water to be 4.18, then yes. I assume you take it as 4.18 everywhere else.

2. Now, you are taking the specific heat capacity of water as 4.32? :confused: If you take 4.18, then it's not right.

3. Yes. Correct.

4. Yes.

5. Correct.

6. Nope.
---------------
I'l show you the 6th one.
1st basic mistake, you have to answer in kJ/mol and not in J.
2nd, you must have been lost in your calculations.

1. Evaluate the heat absorbed by the water.

$Q = mcT = (100)(4.18)(32-23.9) = 3385.8\ J$

2. Now, 3.25 g of NaOH is 0.0825 mol of NaOH (Mr = 40; moles = 3.25/40)

That 0.0825 mol gives off 3385.8 J
Therefore, 1 mole gives off 41041 J (3385.8/0.0825)

That means that $\Delta H = 41041\ J/mol = 41.0\ kJ/mol$

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