# Homework Help: Time to take for water to cool down.

1. Apr 13, 2012

### dzkl

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....

2. Apr 14, 2012

### darkxponent

Have you studied Newtons Law of Cooling?

3. Apr 14, 2012

### dzkl

I saw the mistake...it's suppose to be log base e thus the negative is cancelled out. Btw, ya I've studied Newton's Law

4. Apr 14, 2012

### Staff: Mentor

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. Apr 14, 2012

### darkxponent

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