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

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In summary: I've been over this problem several times and I just can't see where I'm going wrong...In summary, the problem involves cooling 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. The correct final temperature of the water should be 100C, but some students have derived 97C using the formula q(water) = -q(iron). However, this method does not take into consideration the energy required to produce steam. Therefore, the equation q = mcΔT must be used, along with the specific heat capacities of iron (470 J/kgK) and water
  • #1
sdoug041
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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[tex]\Delta[/tex]T 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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  • #2
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[tex]\Delta[/tex]T 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 [tex]Q_{\text{water}}+Q_{\text{iron}}=0[/tex] 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.
 
  • #3
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 [tex]Q_{\text{water}}+Q_{\text{iron}}=0[/tex] 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 [tex]Q_{\text{water}}+Q_{\text{iron}}=0[/tex]? 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?
 
  • #4
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.
 
  • #5
sdoug041 said:
But shouldn't I be yielding a value >100 if I use [tex]Q_{\text{water}}+Q_{\text{iron}}=0[/tex]? 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!
 
  • #6
Alright Well thanks guys I'm just going to assume my prof entered the wrong value for now.
 

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

1. How do you calculate the specific heat capacity of a substance?

The specific heat capacity of a substance can be calculated by dividing the heat energy required to raise the temperature of a given mass of the substance by the mass of the substance and the change in temperature. The formula is: specific heat capacity = heat energy / (mass x change in temperature).

2. What is the heat capacity of water?

The heat capacity of water is 4.184 J/g·°C. This means that it takes 4.184 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

3. How do you calculate the heat transfer in a calorimetry problem?

The heat transfer in a calorimetry problem can be calculated by multiplying the specific heat capacity of the substance by its mass and the change in temperature. The formula is: heat transfer = specific heat capacity x mass x change in temperature.

4. What is the specific heat capacity of iron?

The specific heat capacity of iron is 0.449 J/g·°C. This means that it takes 0.449 joules of energy to raise the temperature of 1 gram of iron by 1 degree Celsius.

5. How do you determine the final temperature in a calorimetry problem?

The final temperature in a calorimetry problem can be determined by using the formula: final temperature = (heat transfer / (specific heat capacity x mass)) + initial temperature. This formula takes into account the heat transfer, specific heat capacities, and initial temperatures of both substances involved in the problem.

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