Cooling Off (Heat, Temperature and Phase Changes)

In summary, the conversation discusses a physics problem where a student tries to cool down a swimming pool by adding ice cubes. The student continues to add ice cubes until the temperature stabilizes to 16oC. The conversation includes the specific heat and latent heat of fusion of water, and provides a formula for calculating the number of ice cubes needed. The student encounters a mistake in their calculation, but ultimately finds the correct answer of 621 ice cubes needed to cool the pool to 16oC.
  • #1
Pat2666
33
0
Okay, so I've approached this problem from a number of different ways and have gotten some crazy answers. The poor student would be spending the day in a bath of ice if my answers were right :P

Here's the problem :

Trying to beat the heat of the last summer, a physics student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, she filled it with 200 liters of water at 25oC. Realizing that the water would probably not be cool enough, she threw ice cubes from her refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0oC.) She continued to add ice cubes, until the temperature stabilized to 16oC. She then got in the pool.
The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g C, the specific heat of ice is 0.5 cal/g C, and the latent heat of fusion of water is 80 cal/g.


How many ice cubes did she add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

Number of Ice Cubes = cubes

HELP: Heat lost by water = heat gained by ice cubes. No heat is lost to the surroundings.

HELP: Since the water (subsystem 1) is at a higher temperature, heat will be lost to the ice cubes (subsystem 2). Calculate the heat H that the water gave up from 25 C to 16 C; calculate the heat h that each ice cube gained from 0C to 16 C including melting. Then the number of ice cubes equals H/h.


So I tried to follow what was given in the help section, determining the Heat Lost by the water and that gained by an ice cube. I then set them up with Qw / Qice as suggested but keeping getting the wrong answer. Any help would be appriciated!

My Work :

http://img206.imageshack.us/img206/620/workjs0.jpg

BTW - I used the standard values for specific heat and latent heat, but I don't believe that should make much of a difference.
 
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  • #2
Pat2666 said:
Okay, so I've approached this problem from a number of different ways and have gotten some crazy answers. The poor student would be spending the day in a bath of ice if my answers were right :P

Here's the problem :

Trying to beat the heat of the last summer, a physics student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, she filled it with 200 liters of water at 25oC. Realizing that the water would probably not be cool enough, she threw ice cubes from her refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0oC.) She continued to add ice cubes, until the temperature stabilized to 16oC. She then got in the pool.
The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g C, the specific heat of ice is 0.5 cal/g C, and the latent heat of fusion of water is 80 cal/g.


How many ice cubes did she add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

Number of Ice Cubes = cubes

HELP: Heat lost by water = heat gained by ice cubes. No heat is lost to the surroundings.

HELP: Since the water (subsystem 1) is at a higher temperature, heat will be lost to the ice cubes (subsystem 2). Calculate the heat H that the water gave up from 25 C to 16 C; calculate the heat h that each ice cube gained from 0C to 16 C including melting. Then the number of ice cubes equals H/h.


So I tried to follow what was given in the help section, determining the Heat Lost by the water and that gained by an ice cube. I then set them up with Qw / Qice as suggested but keeping getting the wrong answer. Any help would be appriciated!

My Work :

http://img206.imageshack.us/img206/620/workjs0.jpg

BTW - I used the standard values for specific heat and latent heat, but I don't believe that should make much of a difference.
What happens to the ice when it has melted? Does it stay at 0oC?
 
Last edited by a moderator:
  • #3
Oh, it continues to heat up?

So I need to add 0.03kg(4186)(16-0) to the demoninator? But it still comes out to the wrong answer. I get 625 ice cubes.

I did 7534800 / (10050 + 2009.28) <--- Original demoniator plus the energy from continued heating.
 
  • #4
Oh I figure out my mistake, I wasn't doing 0.03kg * 2090 because I thought since the change in T was zero it wasn't significant. I guess the student really is going to spend the time adding 621 ice cubes o_0

Thanks for all the help! :)
 

1. How does temperature affect the cooling off process?

The rate at which an object cools off is directly proportional to the difference in temperature between the object and its surroundings. This means that the higher the temperature difference, the faster the object will cool off.

2. What is the difference between heat and temperature?

Heat is the total amount of energy present in an object, while temperature is a measure of the average kinetic energy of the particles in an object. In other words, heat is a form of energy, while temperature is a measure of how hot or cold something is.

3. How do different materials cool off at different rates?

The rate at which an object cools off also depends on its thermal conductivity, which is the ability of a material to transfer heat. Materials with high thermal conductivity, such as metals, cool off faster than materials with low thermal conductivity, such as wood or plastic.

4. What happens to the particles in a substance when it changes phase?

When a substance changes from a solid to a liquid or from a liquid to a gas, the particles in the substance gain energy and move faster, causing the substance to expand. When a substance changes from a gas to a liquid or from a liquid to a solid, the particles lose energy and slow down, causing the substance to contract.

5. Can an object cool off below the temperature of its surroundings?

Yes, an object can cool off below the temperature of its surroundings through a process called evaporative cooling. This occurs when the most energetic particles in a liquid escape into the surrounding air, leaving behind the less energetic particles, resulting in a decrease in temperature.

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