Time to take for water to cool down.

In summary, the conversation is about calculating the time it takes for water in a well-insulated cylindrical cup to cool down from 84degCel to 46degCel. The specific heat capacity of water, U-value, and surface area are provided. The formula used is T = -1/x log (y1-y3)/(y2-y3), with x representing (U-value)(Surface Area)/(mass)(Heat Capacity). The mistake of using the wrong logarithm was pointed out and it was clarified that the argument for the log should be less than one. The conversation ends with the acknowledgement that the speaker has studied Newton's Law of Cooling.
  • #1
dzkl
3
0
My question refers to the cooling of water in a well-insulated , cylindrical cup that is opened at the top.

0.336Kg of water at 84degCel is poured into the cup which is left to stand on a table at an ambient temperature of 25degCel. The specific heat capacity of water is 4150 J Kg^-1 K^-1 and the U-value is 10.64Wm^-2K^-1. The area of the water surface that is exposed to the surroundings air is 0.00447 m^2.

So, how to calculate the time taken in seconds for the water to cool down to 46degCel?

This is what I have done so far...

Let x = (U-value)(Surface Area)/(mass)(Heat Capacity)
Let y1 = 46DegCel (cool down temp)
Let y2 = 84DegCel (Water temp)
Let y3 = 25DegCel (Ambient tempt)
Taking time to be T, so...

T = -1/x log (y1-y3)/(y2-y3)
= -13153 (sec)

I got negative, so my guess something is wrong here. The formula is from my book. Need help...
 
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  • #2
Have you studied Newtons Law of Cooling?
 
  • #3
I saw the mistake...it's suppose to be log base e thus the negative is canceled out. Btw, you I've studied Newton's Law
 
  • #4
dzkl said:
My question refers to the cooling of water in a well-insulated , cylindrical cup that is opened at the top.

0.336Kg of water at 84degCel is poured into the cup which is left to stand on a table at an ambient temperature of 25degCel. The specific heat capacity of water is 4150 J Kg^-1 K^-1 and the U-value is 10.64Wm^-2K^-1. The area of the water surface that is exposed to the surroundings air is 0.00447 m^2.

So, how to calculate the time taken in seconds for the water to cool down to 46degCel?

This is what I have done so far...

Let x = (U-value)(Surface Area)/(mass)(Heat Capacity)
Let y1 = 46DegCel (cool down temp)
Let y2 = 84DegCel (Water temp)
Let y3 = 25DegCel (Ambient tempt)
Taking time to be T, so...

T = -1/x log (y1-y3)/(y2-y3)
= -13153 (sec)

I got negative, so my guess something is wrong here. The formula is from my book. Need help...

Couple of things. First, the log should be the natural logarithm, loge not log10. Second, your argument for the log is less than one, so it should yield a negative value that is subsequently dealt with bu the "-" in front of 1/x.
 
  • #5
dzkl said:
I saw the mistake...it's suppose to be log base e thus the negative is canceled out. Btw, you I've studied Newton's Law

sorry i wasnt able see Newtons law of cooling equation. You have written it in such a way that i thought you have written some other thing. I saw it now
 

Related to Time to take for water to cool down.

1) How long does it take for water to cool down?

The time it takes for water to cool down depends on various factors such as the initial temperature of the water, the temperature of the surrounding environment, and the volume of water. Generally, it takes around 10-15 minutes for a cup of hot water to cool down to room temperature.

2) Does the type of container affect the time it takes for water to cool down?

Yes, the type of container can affect the time it takes for water to cool down. A metal container conducts heat more quickly than a plastic or glass container, so the water in a metal container will cool down faster.

3) Can stirring or shaking the water help it cool down faster?

Yes, stirring or shaking the water can help it cool down faster. This is because it helps distribute the heat evenly throughout the water, making the cooling process more efficient.

4) Does the temperature of the surrounding environment affect the cooling rate of water?

Yes, the temperature of the surrounding environment can greatly affect the cooling rate of water. A higher surrounding temperature will cause the water to cool down faster, while a lower surrounding temperature will slow down the cooling process.

5) Is there a way to speed up the cooling process of water?

Yes, there are a few ways to speed up the cooling process of water. One way is to place the container of water in a larger container filled with ice or cold water. Another way is to add a metal spoon to the water, which will act as a heat sink and help the water cool down faster.

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