Calorimetry Problem: Cooling 90.0 g of Iron in 85.0 g of Water

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Homework Help Overview

The problem involves cooling a 90.0 g slug of iron at 745 °C by placing it into 85.0 g of water at 20.0 °C, within an insulated system. The goal is to determine the final temperature of the water, considering no heat exchange with the surroundings.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss using the equation q(water) = -q(iron) to find the final temperature, but some express confusion over their results, particularly regarding the expected outcome of 100 °C for the water. There is mention of the specific heat capacities for iron and water, and the need to consider the energy required for phase change if the temperature exceeds 100 °C.

Discussion Status

Multiple interpretations of the problem are being explored, with some participants questioning the assumptions made in their calculations. There is a recognition that the energy required for steam production may not have been accounted for in the initial equations. Guidance has been offered regarding the need to adjust calculations based on the next part of the question.

Contextual Notes

Participants note that the specific heat of iron is assumed to be constant and not temperature dependent. There is also a concern regarding the implications of the calculated temperature being below or above 100 °C in relation to the phase change of water.

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Homework Statement



You cool a 90.0 g slug of red-hot iron (temperature 745 C) by dropping it into an insulated cup of negligible mass containing 85.0 g of water at 20.0 C. Assume no heat exchange with the surroundings.

What is the final temperature of the water?


Homework Equations



q = mc\DeltaT and most likely q = mL



The Attempt at a Solution



I know this is a fairly straight forward question but I am not receiving the right value for the temperature of the water. I use the formula q(water) = -q(iron) and solve for T(2). I keep getting back 97C, but the correct answer (apparently) should be 100C because for the next part of the question you need to figure out how much water is left over from some of it converting to steam.

Here is some additional information: The specific heat capacity I'm using for Iron is 470 J/kgK and for water 4190 J/kgK
 
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sdoug041 said:

Homework Statement



You cool a 90.0 g slug of red-hot iron (temperature 745 C) by dropping it into an insulated cup of negligible mass containing 85.0 g of water at 20.0 C. Assume no heat exchange with the surroundings.

What is the final temperature of the water?

Homework Equations



q = mc\DeltaT and most likely q = mL

The Attempt at a Solution



I know this is a fairly straight forward question but I am not receiving the right value for the temperature of the water. I use the formula q(water) = -q(iron) and solve for T(2). I keep getting back 97C, but the correct answer (apparently) should be 100C because for the next part of the question you need to figure out how much water is left over from some of it converting to steam.

Here is some additional information: The specific heat capacity I'm using for Iron is 470 J/kgK and for water 4190 J/kgK
I'd be glad to see your derivation of 97°C.
I suggest you to use some intuition, or as you did, reading the next part of the question. It is clear that water will reach 100 °C and a portion of it will steam.
So writing down Q_{\text{water}}+Q_{\text{iron}}=0 is not appropriated, because you don't take in count the energy required to steam up the water.
You have to add a term in the equation, the one you wrote under the "Revelant equations".
Good luck!
Edit: You could also do some research on the forum. There are many, many similar problems in the database.
 
fluidistic said:
I'd be glad to see your derivation of 97°C.
I suggest you to use some intuition, or as you did, reading the next part of the question. It is clear that water will reach 100 °C and a portion of it will steam.
So writing down Q_{\text{water}}+Q_{\text{iron}}=0 is not appropriated, because you don't take in count the energy required to steam up the water.
You have to add a term in the equation, the one you wrote under the "Revelant equations".
Good luck!
Edit: You could also do some research on the forum. There are many, many similar problems in the database.

But shouldn't I be yielding a value >100 if I use Q_{\text{water}}+Q_{\text{iron}}=0? You can't just know that it's going to be greater than 100 then add the additional term... Aren't you supposed to test that by using the equation to see if you get over 100? And then if you do you need to take in consideration the thermal energy used to produce the steam?
 
I got 97 deg C as well.

That is assuming specific heat of iron is 470 J/kgK and is not temperature dependent.

Edit: and I wouldn't worry about steam. As long as the answer is BELOW 100 deg C, we know we are on the safe side. Once the answer becomes 103 deg C, we know our equation for heat balance was wrong.

Edit 2: that's exactly what you wrote - and you are right about it.
 
sdoug041 said:
But shouldn't I be yielding a value >100 if I use Q_{\text{water}}+Q_{\text{iron}}=0? You can't just know that it's going to be greater than 100 then add the additional term... Aren't you supposed to test that by using the equation to see if you get over 100? And then if you do you need to take in consideration the thermal energy used to produce the steam?
You're right. That's why I said I'd glad to see your derivation.
Borek said:
I got 97 deg C as well.

That is assuming specific heat of iron is 470 J/kgK and is not temperature dependent.

Edit: and I wouldn't worry about steam. As long as the answer is BELOW 100 deg C, we know we are on the safe side. Once the answer becomes 103 deg C, we know our equation for heat balance was wrong.

Edit 2: that's exactly what you wrote - and you are right about it.
Oh well, my intuition is dead wrong, once more!
 
Alright Well thanks guys I'm just going to assume my prof entered the wrong value for now.
 

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