Heat transfer: ice in water bath

In summary, a physics grad student bought a plastic child's swimming pool and filled it with 200 liters of water at 25 degrees Celsius. He threw ice cubes from his refrigerator, each of mass 30 grams, into the pool to increase the pool's temperature to 16 degrees Celsius.
  • #1
huskydc
78
0
During one hot summer, a physics grad student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, he filled it with 200 liters of water at 25 C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0C.) He continued to add ice cubes, until the temperature stabilized to 16C. He then got in the pool.


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

ok..i tried doing the following:

heat lost by water= heat gained by ice cubes

Q water = m c delta T

from the fact d=m/v, i got mass of water as 200 kg

so Q water = 200 (4186) 9 = 7534800 J

then I'm not so sure as to where i go from here...

I've been told to find the heat gained by ice cubes including the melting...how should i do that?
 
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  • #2
I'm sure your textbook discusses the heat of fusion for water.
 
  • #3
but the problem is I'm not sure how to set it up...

i try something like this, and i know it's wrong...

Q water =7534800 J

each ice cube is .3 kg,

Q ice = m c delta T: .3(2090)(16) = 10032 J

melting? Q = 200 x (33.5 e 4) = 6.7 e 7


i'm just confused...
 
  • #4
i'm sorry folks, I'm really lost with this one? can someone help me out?
 
  • #5
I'll have to give it some thought. It's unclear to me why they introduce the notion of "hot day" and then don't provide any additional information about the "ambient temperature." Given that, apparently they want you to consider only the melting and heat transfer without regard to the ambient temperature.

I think the basic idea is that the mass of the ice decreases while the mass of liquid water increases. Energy is required to melt the ice (latent heat of fusion) and to warm it up the to the temperature of the liquid water while the liquid water supplies the heat required to do that at a rate proportional to the temperature difference between the liquid water and the ice.

From the statement of the problem, it would appear that at some particular ratio of ice to liquid that the temperature difference is "stationary" (16 C) - that's what I need to ponder!

In the meantime, anyone else should feel free to jump in - i.e. the problem, not the pool!
 
  • #6
Tide said:
I think the basic idea is that the mass of the ice decreases while the mass of liquid water increases. Energy is required to melt the ice (latent heat of fusion) and to warm it up the to the temperature of the liquid water while the liquid water supplies the heat required to do that at a rate proportional to the temperature difference between the liquid water and the ice.
From the statement of the problem, it would appear that at some particular ratio of ice to liquid that the temperature difference is "stationary" (16 C) - that's what I need to ponder!
I think you have to assume that all the ice cubes melt. Otherwise there is no equilibrium and the pool keeps getting colder (assuming no heat is transferred from the surroundings). So:

[tex]\Delta Q_{water} = mc\Delta T_1[/tex]

where [itex]\Delta T_1 = 9 ^oC; m = 200 kg.; c = 1 cal./^oC g = 4.187 kJ./^oC kg.[/itex]

[tex]\Delta Q_{ice} = m_{ice}(h_f + c\Delta T_2)[/itex]

where [itex]\Delta T_2 = 16 ^oC; h_f = 334 kJ/kg. [/itex]

Equate the two heats. Since the only unknown is the ice mass, you should be able to work it out.
AM
 
Last edited:
  • #7
ok...so...

Q water = 200 (4.186)(9) = 7534.8

Q ice = m (334 + (2.093 x 16)) = 367.488 m

Q water = Q ice

7534.8 / 367.488 = m of ice = 20.5 kg...

since each cube weighs .03 kg... 20.5/.03 = 684 cubes?

i tried that didn't work...unless the units I'm working with are wrong?
 
  • #8
huskydc said:
ok...so...
Q water = 200 (4.186)(9) = 7534.8
Q ice = m (334 + (2.093 x 16)) = 367.488 m
Q water = Q ice
7534.8 / 367.488 = m of ice = 20.5 kg...
since each cube weighs .03 kg... 20.5/.03 = 684 cubes?
i tried that didn't work...unless the units I'm working with are wrong?
For Q ice you have to use 4.186 kJ/kg. as the specific heat (the specific heat of water) because it goes from 0 - 16 degrees as water. I get 626 ice cubes.

AM
 
  • #9
ok...that works...i was thinking since it's heat gained by ice cubes...i'll use the specific heat for ice...

then may i ask under what kind of situation do i use specific heat for ice??

conditions where..ice...is still in temp... below zero??
 
  • #10
huskydc said:
then may i ask under what kind of situation do i use specific heat for ice??
conditions where..ice...is still in temp... below zero??
When it is ice and not liquid water.

AM
 

1. How does heat transfer occur between ice and water in a bath?

Heat transfer occurs through a process called conduction, where heat energy is transferred from the warmer substance (water) to the colder substance (ice) through direct contact. The molecules in the warmer substance have higher kinetic energy, causing them to collide with the colder molecules and transfer heat energy.

2. Why does the ice melt in a water bath?

The ice melts because the heat energy from the warmer water is transferred to the ice, causing the ice molecules to gain energy and break apart, transitioning from a solid to a liquid state. This process is called melting.

3. Does the temperature of the water affect the rate of heat transfer to the ice?

Yes, the temperature of the water does affect the rate of heat transfer. The greater the temperature difference between the water and the ice, the faster the heat transfer will occur. This is because there is a greater temperature gradient, allowing for more efficient heat transfer.

4. How does the surface area of the ice affect heat transfer in a water bath?

The surface area of the ice does affect heat transfer. A larger surface area allows for more direct contact between the ice and the water, resulting in a faster rate of heat transfer. This is why crushed or smaller ice cubes will melt faster in a water bath compared to larger ice cubes.

5. Can heat transfer also occur through the process of convection in a water bath?

Yes, heat transfer can also occur through the process of convection in a water bath. Convection is the transfer of heat energy through the movement of fluids (in this case, the water). As the warmer water rises and the cooler water sinks, it creates a convection current that helps distribute the heat energy throughout the water bath, resulting in more efficient heat transfer to the ice.

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