How do you solve calorimetry problems?

[flyingpig]

Homework Statement

A 125-g stainless steel ball bearing at 545C⁰ (steel has a specific heat capacity of 0.50J g⁻¹C⁻¹) is dropped into 80mL of water at 27.5C⁰ in an open Styrofoam cup. As a result, the water is brought to a boil when the temperature reaches 100.0C⁰.

What mass of water vaporizes while the boiling continues? (∆H⁰vap = 40.6kJ/mol H₂O)

The Attempt at a Solution

m_w = -\frac{m_s c_s (100 - 545)}{c_w (100 - 27.5)}

I got 91.7g, but it is wrong

 

[gneill]

I don't see the original mass of the water in your expression, nor where you've accounted for the heat required vaporize at least some of it. Further, since there are only 80 grams of water to begin with (water has a density of 1 gram per mL), it's a miracle that you've managed to evaporate nearly 92 grams of the stuff!

 

[flyingpig]

There is no original mass, this is all I have been given.

 

[gneill]

There is no original mass, this is all I have been given.

"...is dropped into 80mL of water..."

 

[flyingpig]

80g? 1g/mL? What does that have to do with the heat though?

 

[flyingpig]

91.7g - 80g = 11.7g?

 

[gneill]

Yer spitballing. Work though the problem one step at a time.

The steel ball is dropped into the water. No boiling can take place until all the water reaches the boiling point. How much heat does that take? Where does that heat come from?

 

[flyingpig]

Q = (80g)(4.18)(100-27.5) = 24.244J?

What do I do now?

 

[gneill]

Okay, your method's correct but you've lost some orders of magnitude. Check your units. Once you've fixed it...

Where did that heat to raise the temperature of the water come from? What's the new temperature of the source?

 

[flyingpig]

(80g)(4.18J/g C)(100-27.5) = 24.244J? They look fine to me

The heat comes from the ball, the new temperature is 545C.

 

[gneill]

(80g)(4.18J/g C)(100-27.5) = 24.244J? They look fine to me

The heat comes from the ball, the new temperature is 545C.

So, 80 x 4.18 x 72.5 = 24.2? Must be the "new math"!

And 545C was the original temperature of the steel ball.

 

[flyingpig]

*24244J, my calculator put a comma...

*100 degrees, when they reached that temperature is the new temperature.

 

[gneill]

*24244J, my calculator put a comma...

That's better!

*100 degrees, when they reached that temperature is the new temperature.

Okay, we can work it that way. I was going to suggest calculating the new temperature of the steel ball once the water had reached 100C and before any water had started turning to steam. But we can skip over that step and account for it later. Everything must be accounted for in the end!

IF we assume that there will be some water left after the steel ball reaches 100C and the boiling stops, how much heat in total will come from the steel ball and go into water?

 

[flyingpig]

Q = (125g)(0.5)(100-545) = -27812.5J

 

[gneill]

Q = (125g)(0.5)(100-545) = -27812.5J

Okay. That's the total amount of heat available from the steel ball to work its magic on the water from start to finish.

Some of that capital has already been spent on raising the temperature of the water to 100C. What's remaining to turn water to steam?

 

[flyingpig]

My breathe...? I don't know.

 

[gneill]

You've calculated the total heat available from the steel ball. You've calculated the amount of heat it took to raise the water temperature to 100C. What's left of the heat available?

 

[flyingpig]

24244J + -27812.5J = 3568J

The excess heat

40.6kJ/mol H₂O x (80g/18g/mol) = 180444J

Now I am stuck again...

 

[gneill]

Why are you assuming that all the water will be turned to steam?

First convert the ∆H from kJ/mol to kJ/g without worrying about the amount of water involved; get that value first.

 

[flyingpig]

40.6kJ/mol / 18g/mol = 2.25kJ/g

 

[gneill]

Excellent. Now, your energy budget remaining is a paltry 3568 J. You have 3568 Joules available to convert water to steam at a going rate of 2.25 kJ per gram of water converted. How many grams can you afford to convert?

 

[flyingpig]

3568J/2.25kJ/g = 1.585g? Did I get it right!?

 

[flyingpig]

Nope I am right! (well you are right...)

 

[gneill]

3568J/2.25kJ/g = 1.585g? Did I get it right!?

You have reached your destination. Now see if you can repeat the process without looking at these notes!

Working these problems is a matter of keeping track of where the energy is, where it's going, how much is available, and how much required to accomplish the various "steps" (like raising or lowering temperatures to specific points where phase changes occur). It's a lot like being an accountant, where the currency is heat.