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Newton's Law of Cooling: DE 730 MW

The document describes Newton's Law of Cooling and Heating, which states that the rate of temperature change of an object is proportional to the difference between the object's temperature and the temperature of the surrounding medium. The law is applied to several examples of objects cooling or heating in different mediums over time to calculate temperature values at given times. Solutions are provided for examples involving determining the temperature of a thermometer or other object at future times based on its initial temperature, the temperature of the surrounding medium, and temperature measurements at other times during the cooling or heating process.

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Kim Chua
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386 views2 pages

Newton's Law of Cooling: DE 730 MW

The document describes Newton's Law of Cooling and Heating, which states that the rate of temperature change of an object is proportional to the difference between the object's temperature and the temperature of the surrounding medium. The law is applied to several examples of objects cooling or heating in different mediums over time to calculate temperature values at given times. Solutions are provided for examples involving determining the temperature of a thermometer or other object at future times based on its initial temperature, the temperature of the surrounding medium, and temperature measurements at other times during the cooling or heating process.

Uploaded by

Kim Chua
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOC, PDF, TXT or read online on Scribd
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DE 730 MW

Newton’s Law of Cooling

For cooling process For heating process


dT dT
  k  T  Tm   k  Tm  T 
dt dt
where : T  temperature of the body
t  time
Tm  temperature of the medium
k  cons tan t of proportionality

We assume that the rate of cooling is equal to the rate of heating. (k cooling = kheating)

Example
A thermometer reading is 18ºF is brought into a room where the temperature is 70ºF; a minute
later, the thermometer reading is 31ºF. Determine temperature reading 5 minutes after the
thermometer is first brought into the room. 58ºF

dT
 k  Tm  T 
dt
dT
 k  70  T 
dt
dT
 kdt
 70  T 
dT
  k dt   0
 70  T  
 ln  70  T   kt  c
BC1: t  0 T  18
 ln  70  18   k  0   c
c   ln52
BC2 : t  1 T  31
 ln  70  31  k  1   ln52
k  0.2877 / min
BC3 : t  5 T?
 ln  70  T   0.2877  5    ln52
T  57.66 F

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DE 730 MW
Newton’s Law of Cooling/Heating

1. The temperature of air is 30ºC, and the substance cools from 100ºC to 70ºC in 15 minutes.
Find t when the temperature will be 40ºC. 52.20 min.

2. At 1:00 PM, a thermometer reading 70ºF is taken outside where the air temperature is –10ºF.
At 1:02 PM, the reading is 26ºF. At 1:05 PM, the thermometer is taken back indoors, where
the air is 70ºF. What is the temperature reading at 1:09 PM? 56ºF

3. At 9:00 AM, a thermometer reading 70ºF is taken outdoors, where the temperature is 15ºF.
At 9:05 AM, the thermometer reading is 45ºF. At 9:10 AM, the thermometer is taken back
indoors, where the temperature is fixed at 70ºF. Find the reading at 9:20 AM.
58.5ºF
4. At 2:00 PM, a thermometer reading 80ºF is taken outside where the air temperature is 20ºF.
At 2:03 PM, the temperature reading yielded by the thermometer is 42ºF. Later, the
thermometer is brought inside, where the air is at 80ºF. At 2:10 PM, the reading is 71ºF.
When was the thermometer brought indoors? 2:05 PM

5. An 8-lb metal has a specific heat of 1/32, while at a temperature of 308ºF. It is dropped into
11 lb of water of which the temperature is 53ºF. Find T at t = 16 min, if T = 200ºF at t =10
min.
159.2ºF

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