Calculating Heat Absorbed & Temperature Change of Rocks

In summary: Well okay, this is what I got when I tried to solve for delta-T (I tried twice and did something wrong, I can't remember how to isolate it because it is in the denominator...) the -x- represents that variable being cancelled.ΔT=(0.82 J/g-K)(50.0g)/450kJThat answer was far from correct. I also tried multiplying all my given variables, also far from correct :(
  • #1
FTCC.student
7
0

Homework Statement


Large beds of rock are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. (a) Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0°C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?

Homework Equations



Cs=[itex]\frac{q}{m * ΔT}[/itex]

The Attempt at a Solution



For part a, I have rearranged the formula to give me q=Cs X m X ΔT, and got an answer of 4.9*105 J. Part B, however, all I could think to do was try and rearrange the equation again by solving for ΔT. Either I cannot figure out how to rearrange it (this is probably the case, I was never very good at that), or I am on the wrong track completely. Either way, I'm probably also over-thinking it.

Also, a question about forum rules, I am studying for a test that is tomorrow afternoon and will probably have several other questions tonight on a variety of subjects (such as ΔH, electron config, net ionic equations, concentrations of solutions/molarity etc). Am I allowed to create a new thread for each question or should I stick to this thread?

Thanks for the help.
 
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  • #2
Hi FTCC.student! :smile:
FTCC.student said:
(b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?

Part B, however, all I could think to do was try and rearrange the equation again by solving for ΔT.

Should work.

Show us how far you got.
… Am I allowed to create a new thread for each question …

yes :smile:

quick! :wink:
 
  • #3
Well okay, this is what I got when I tried to solve for delta-T (I tried twice and did something wrong, I can't remember how to isolate it because it is in the denominator...) the -x- represents that variable being cancelled.


ΔT = [itex]\frac{Cs*m}{q}[/itex]

Which gave me: ΔT=(0.82 J/g-K)(50.0g)/450kJ

That answer was far from correct. I also tried multiplying all my given variables, also far from correct :(

So what I'm really confused on here probably isn't the actual chemistry portion, just how to rearrange the dang formula
 
  • #4
FTCC.student said:
ΔT = [itex]\frac{C_s*m}{q}[/itex]
(Your LaTex was messed up because you used "" inside [itex]. That doesn't work. You have to use _ for subscript, ^ for superscript, etc.)
Let's try that one step at a time:
Cs=[itex]\frac{q}{m * ΔT}[/itex]​
Multiply both sides by ΔT
ΔT*Cs=[itex]\frac{q}{m}[/itex]​
What next?
 
  • #5


Hello,

I would like to provide a response to your content about calculating heat absorbed and temperature change of rocks.

For part a, your approach to solving for the quantity of heat absorbed by the rocks is correct. However, I would recommend using the specific heat capacity in its unit of J/kg-K instead of J/g-K, as this will give you a more accurate answer. Thus, the formula would be q = Cs * m * ΔT, where Cs = 0.82 J/kg-K. This would give you an answer of 4.1*10^6 J, which is equivalent to 4.1 MJ.

For part b, you are on the right track by rearranging the formula to solve for ΔT. The correct formula would be ΔT = q/(Cs * m). Plugging in the values of q = 450 kJ, Cs = 0.82 J/kg-K, and m = 50.0 kg, we get a temperature change of 0.109°C.

In terms of forum rules, it is generally recommended to create a new thread for each question to keep the discussion organized. However, if your questions are related to the same subject, it may be better to keep them in the same thread. I hope this helps and good luck on your test tomorrow!
 

1. How do I calculate the heat absorbed by a rock?

To calculate the heat absorbed by a rock, you can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the rock, c is the specific heat capacity of the rock, and ΔT is the change in temperature. You will need to know the initial and final temperatures of the rock, as well as the specific heat capacity of the type of rock you are working with.

2. What is specific heat capacity?

Specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius. It is a measure of how much energy a substance can absorb before its temperature changes.

3. How do I determine the change in temperature of a rock?

To determine the change in temperature of a rock, you will need to measure its initial temperature and its final temperature after it has absorbed heat. The change in temperature, or ΔT, can then be used in the formula Q = mcΔT to calculate the heat absorbed by the rock.

4. Can I use the same specific heat capacity for all types of rocks?

No, specific heat capacity varies depending on the type of rock. Different types of rocks have different compositions and densities, which affect their ability to absorb heat. It is important to use the correct specific heat capacity value for the specific type of rock you are working with.

5. What are some factors that can affect the accuracy of my calculations for heat absorbed and temperature change of rocks?

Some factors that can affect the accuracy of your calculations include variations in the specific heat capacity of the rock, errors in measurement of initial and final temperatures, and heat loss to the surrounding environment. It is important to carefully control these factors and use accurate data in your calculations to ensure the most precise results.

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