EK 1001 Chemistry: Question pls help! What's the answer and why?

[xyz11]

In careful calculations with calorimeters, the heat capacity of such items as the thermometer and the stirring rod must be considered. Suppose a coffee-cup calorimeter is used to calculate the heat of solution for sodium hydroxide, but the experimenter does not include the heat capacity of the thermometer and stirrer. How would this affect the results?


A. The calculated value of the heat of solution would be too low
B. The calculated value of the heat of solution would be too high
C. The calculated value of the heat of solution would be unaffected
D. The effect on the results depends on whether the reaction is endothermic or exothermic

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[BennieBlanco]

In careful calculations with calorimeters, the heat capacity of such items as the thermometer and the stirring rod must be considered. Suppose a coffee-cup calorimeter is used to calculate the heat of solution for sodium hydroxide, but the experimenter does not include the heat capacity of the thermometer and stirrer. How would this affect the results?


A. The calculated value of the heat of solution would be too low
B. The calculated value of the heat of solution would be too high
C. The calculated value of the heat of solution would be unaffected
D. The effect on the results depends on whether the reaction is endothermic or exothermic

Are you sure this is EK1001? I just scanned it and couldn't find this question in thermochem.

Anyway. consider an exothermic reaction...

Heat is lost by the reaction (reactants/pdts) and gained by the solution. But in reality some of the heat will go to the thermometer and the stirrer.

Therefore, less heat will go to the solution (making the temperature change smaller than it should be). You would expect the specific heat to be higher than it actually is.

Clarify this by examining what specific heat is: s = q / (m x dT)

if the temperature change is smaller than it should be then the s value will increase.

 

[xyz11]

Hi BennieBlanco,

Thanks for replying to this. The prob was #384 on EK1001. The answer they have is A, my TA said that it's D, so I was confused. Your explanation definitely clears things up, though! Thanks so much! 🙂

 

[BennieBlanco]

Hi BennieBlanco,

Thanks for replying to this. The prob was #384 on EK1001. The answer they have is A, my TA said that it's D, so I was confused. Your explanation definitely clears things up, though! Thanks so much! 🙂

I screwed up my reasoning thinking it was B, but it is A.

So I previously stated that some of the heat would be absorbed by the thermometer/stirrer, I concluded that the solution would have less of a temperature change, which would result in a higher specific heat (actually the reasoning is correct but the conclusion was incorrect).

In actuality the thermometer and the stirrer also RESIST temperature change (i.e. they must be warmed up). Because they must be warmed up, they too have specific heats. Once you total the dT from the reaction, you are assuming all the specific heat came from the solution when in reality it came from both the solution and the thermometer/stirrer.

So in other words, these contribute to heat resistance (actual): solution/therm/stirrer.

The calculations would only account for the solution. Therefore the answer is A.

Applying arbitrary #'s:

Solution: 3
Thermometer: 0.5
Stirrer: 0.5

Total specific heat: 4

Actual specific heat (solution): 3

 

[MLT2MT2DO]

Wouldn't your calculated be the 4.0 (the heat required for the reaction) even though the heat of the solution was 3.0? Making the answer "B"?

 

[BennieBlanco]

Wouldn't your calculated be the 4.0 (the heat required for the reaction) even though the heat of the solution was 3.0? Making the answer "B"?

Here is a better explanation. By the way, I used the term "specific heat" repeatedly because I did some thermo reading yesterday. We are actually referring to heat capacity, which is the molar equivalent.

Use the equation q = mc dT

If we think of the heat capacity, then it is true that it would be higher than we expected. But in this problem the are asking us to solve for q. They are not asking us to solve for C. They are assuming C was a value we looked up in a book. So back to our arbitrary #'s, we decided C for the solution was 3. Lets also arbitrarily decide the temperature change dT is 2.

q = 3 x m x 2 = 6m (calculated) TOO LOW

Well, we would then conclude the heat of the reaction was 6 x the mass.

The reality would be that it would be:

q = 4 x m x 2 = 8m (actual) HIGHER

So I think the confusion stems from thinking you are calculating the heat capacity (C), when it actuality you are calculating the heat of solution (q). This is why I answered incorrectly at first, read the Q wrong. If you read my first response, the reasoning is correct but it is attempting to answer the WRONG question... don't read it, it will confuse (although conceptually ok)

 

[HalfPast6]

Was trying to wrap my brain around this solution until I reinterpreted the question. Is this an appropriate way of coming up with the solution?

Question is basically saying that you are not accounting for C of the stirrer and thermometer, then the C used in your formula is lower than expected. If you know that q=C x dT & dq=dH (heat of solution), then q will also be lower. Therefore, H will be lower.