# Sear's and Zemansky Ex. 17.10: What's Cooking?

• FountainDew
In summary, a heavy copper pot at 150°C has 0.10 kg of water at 25°C added to it, and the lid is quickly closed to prevent any steam from escaping. The final temperature of the pot and water is 100°C, and 3.4 grams of the water has been converted to steam due to the energy lost by the copper pot. This conversion was calculated using the heat transfer equation, assuming conservation of energy.
FountainDew

## Homework Statement

A heavy copper pot of mass 2.0 kg (including the copper lid) is at a temperature of 150°C. You pour 0.10 kg of water at 25°C into the pot, then quickly close the lid of the pot so that no steam can escape. Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings.

m(copper) = 2.0 kg
T(copper) = 150°C
c(copper) = 390 J/kg*K

m(water) = 0.10 kg
T(water) = 25°C
c(water) = 4190 J/kg*K
Lv(water) = 2256x10^3 J/kg

## Homework Equations

Heat transfer: Q = mc(delta)T
Heat transfer in Phase Change: Q = m*(Lv)
Conservation of energy: Q(pot) + Q(water) = 0

## The Attempt at a Solution

I calculated the final T to be 106°C for the pot/water system, that was the easy part. The next part says the real final Temp is 100°C because the energy that would have gone into heating the water, actually went into changing its phase from liquid to gas. I don't understand how to do this part at all. I tried calculating a second Q for a (delta)T of 6°C and got a value in Joules. Then I plugged this energy into Q = m*(Lv) and got a mass of 1.1g water vapor which is wrong. So my approach is flawed, why?

4. The actual solution
The final temperature is 100°C, and of the 0.10kg of water in the pot, 0.0034kg or 3.4g of the water was changed into vapor.

The "heat of vaporization" for water is 2260000 J/kg according to
http://en.wikipedia.org/wiki/Latent_heat
I guess modify the first stage of your calc to just go up to 100 degrees and see how much heat is leftover. Put that into heat of fusion = m*2260000 J/kg and see what mass of water is vaporized.

Kind of contrived: if the lid prevents steam from escaping, the pressure will go up and the boiling temp will rise . . . so quite possibly your first answer is better than the one in the back of the book!

Eureka!

5. The actual solution... again!
The solution was simple. I applied the heat transfer equation Q = mc(delta)T for both the water and the copper pot. I assumed the final temperature to be 100°C and calculated both values separately. Since energy is conserved, I knew that Q(pot) + Q(water) = 0 but then I immediately saw there was a difference that did NOT equal 0. That difference is the energy the copper pot lost to convert water into steam! In my case it was -7575 J, then it was a matter of figuring out how much steam was converted using Q = m*Lv. Regardless of which Lv I used for water, the magic of sig figs resulted in a m = 0.0034 kg or 3.4g! Turns out the book was right!

## 1. What is Sear's and Zemansky Ex. 17.10: What's Cooking?

Sear's and Zemansky Ex. 17.10: What's Cooking is a physics problem from the textbook "University Physics" by Sears and Zemansky. It involves calculating the amount of heat needed to cook a meal using principles of thermodynamics.

## 2. What are the main concepts involved in this problem?

The main concepts involved in this problem are heat transfer, specific heat capacity, and the first law of thermodynamics. Heat transfer is the movement of thermal energy from one object to another. Specific heat capacity is the amount of heat required to raise the temperature of a substance by 1 degree Celsius. The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted.

## 3. How do you approach solving this problem?

To solve this problem, you must first identify the given information, such as the mass and initial and final temperatures of the meal. Then, you can use the equation Q = mcΔT (where Q is heat, m is mass, c is specific heat capacity, and ΔT is change in temperature) to calculate the amount of heat needed to raise the temperature of the meal. Finally, you can use the first law of thermodynamics to determine the source of the heat, such as a stove or oven.

## 4. Can this problem be applied to real-life situations?

Yes, this problem can be applied to real-life situations. Understanding the principles of heat transfer and thermodynamics is essential in cooking and food preparation. By calculating the amount of heat needed to cook a meal, you can adjust cooking times and temperatures to achieve the desired result.

## 5. Are there any other resources available for further understanding of this problem?

Yes, there are many online resources available for further understanding of this problem, including videos, practice problems, and tutorials. Additionally, consulting a physics textbook or seeking help from a physics tutor can also aid in understanding this problem and its concepts.

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