In the previous section we saw heat transfer by convection and radiation. In this section, we will see Newton's Law of cooling
Cooling is same as 'loss of heat'. Some basics about loss of heat can be written in 4 steps:
1. Consider a cup of tea placed on the table
• We know that, as time passes, the tea will become lesser and lesser hot
♦ This is because, heat flows from the tea into the surroundings
♦ That means, the tea loses heat
2. We want to know the rate at which heat is lost
‘Rate at which heat is lost’ can be explained in 4 steps:
(i) Let us record the temperature of the tea at various instances using a thermometer and a stop-watch
• Let the temperature of the tea be:
♦ T1 when the reading in the stop-watch is t1
♦ T2 when the reading in the stop-watch is t2
(ii) Then the differences can be calculated as:
♦ difference in temperature = (T1 - T2)
♦ difference in time = (t2-t1)
(iii) The difference in temperature (T1 - T2) can be related to ‘difference in heat contents’ using the specific heat capacity of the tea
• The specific heat capacity is a constant. So the difference (T1- T2) will give an indication about the ‘heat loss’ during the time interval (t2- t1)
(iv) So the rate at which heat is lost will be equal to $\mathbf\small{\rm{\frac{T_1-T_2}{t_2-t_1}}}$
3. But to confirm the result in 2(iv), we have to repeat the experiment
• We will repeat the experiment on the same tea
• The tea is continuously loosing heat. We will take one more set of readings:
(i) Let the temperature of the tea be:
♦ T3 when the reading in the stop-watch is t3
♦ T4 when the reading in the stop-watch is t4
(ii) Then the differences can be calculated as:
♦ difference in temperature = (T3 - T4)
♦ difference in time = (t4 - t3)
(iii) As before, the difference (T3- T4) will give an indication about the ‘heat loss’ during the time interval (t4 - t3)
(iv) So the rate at which heat is lost will be equal to $\mathbf\small{\rm{\frac{T_3-T_4}{t_4-t_3}}}$
4. Here we see a problem:
• The result in 2(iv) will not be equal to result in 3(iv)
• That means:
The tea is losing heat at different rates
• Such a situation needs further investigation. So we will do an experiment in the lab
The experiment can be explained in 22 steps:
1. Take 300 mL water in a calorimeter
• Cover it with a two holed lid
♦ Put a stirrer through one hole
♦ Put a thermometer through the other hole
✰ Make sure that, the bulb of the thermometer is immersed in water
2. Note the reading of the thermometer
♦ This reading will be the temperature of the surroundings
♦ We will denote it as Ts
3. Heat the water until it’s temperature becomes about (Ts + 40) °C
• Stop heating by removing the heat source
4. Start the stop-watch
• The initial reading in the stop-watch will be 0 s
♦ We will denote this instant as t(0)
• At that instant t(0), take the reading in the thermometer
♦ Let it be T(0)
• So at the instant t(0), the water is at a temperature of (T(0)-Ts) higher than the surrounding temperature
5. Stir the water gently using the stirrer
Continue stirring until the reading in the stop-watch is 60 s
♦ We will denote this instant as t(60)
• At that instant t(60), take the reading in the thermometer
♦ Let it be T(60)
• So at the instant t(60), the water is at a temperature of (T(60)-Ts) higher than the surrounding temperature
6. Continue stirring until the reading in the stop-watch is 120 s
♦ We will denote this instant as t(120)
• At that instant t(120), take the reading in the thermometer
♦ Let it be T(120)
• So at the instant t(120), the water is at a temperature of (T(120)-Ts) higher than the surrounding temperature
7. Continue stirring until the reading in the stop-watch is 180 s
♦ We will denote this instant as t(180)
• At that instant t(180), take the reading in the thermometer
♦ Let it be T(180)
• So at the instant t(180), the water is at a temperature of (T(180)-Ts) higher than the surrounding temperature
8. As the water is losing heat, T(0), T(60), T(120) . . . etc., will be progressively decreasing
• So (T(0)-Ts), (T(60)-Ts), (T(120)-Ts) . . . etc., will be progressively decreasing
9. Continue the procedure a total of n times until (T(n)-Ts) is about 5 °C
• That means, the last reading is taken at the instant when the water is about 5 °C above the surrounding temperature
♦ Recall that, at the first instant, the difference (T(0)-Ts) was about 40 °C
10. We can tabulate the readings as shown below:
11. Now we can plot a graph
♦ The values in the first column of the table are plotted along the x-axis
♦ The values in the third column of the table are plotted along the y-axis
• The result is the red curve shown in fig.11.20(a) below:
 |
| Fig.11.20 |
• The following steps from (12) to (22) will give us some interesting information about the above graph:
12. On the x-axis, mark any four points P, Q, U and V in such a way that, PQ = UV
♦ This is shown in fig.11.20(b)
♦ Since PQ is equal to UV, we are taking two ‘equal time duration’
13. Draw vertical green dashed lines through those four points
♦ The vertical dashed lines will intersect the red curve
14. Through those points of intersection, draw horizontal green dashed lines
♦ The horizontal dashed lines will intersect the y-axis at P’, Q’, U’ and V’
15. Consider the distance between P’ and Q’
• P’ is the temperature: T(P) – Ts
♦ It is the temperature of the water (above surrounding temperature) when time is P
• Q’ is the temperature: T(Q) – Ts
♦ It is the temperature of the water (above surrounding temperature) when time is Q
• So we get:
Distance between P’ and Q’
= [(T(P) – Ts) - (T(Q) – Ts)] = [T(P) – T(Q)]
• So P’Q’ indicates the heat lost in the duration PQ
16. Similarly, we can prove:
U’V’ indicates the heat lost in the duration UV
17. The duration PQ and duration UV are the same
• But we see that, the heat loss U’V’ is far lass than the heat loss P’Q’
18. We can write in the terms of rate at which heat is lost:
• Rate at which heat is lost during PQ = $\mathbf\small{\rm{\frac{P'Q'}{PQ}}}$
• 'Rate at which heat is lost' is same as 'Rate of cooling'
• So we get:
Rate of cooling during PQ = $\mathbf\small{\rm{\frac{P'Q'}{PQ}}}$
19. Similarly, we get:
Rate of cooling during UV = $\mathbf\small{\rm{\frac{U'V'}{UV}}}$
20. The denominators in (18) and (19) are the same
• The numerator in (18) is larger than the numerator in (19)
♦ So the result in (18) will be larger than the result in (19)
• So we can write:
♦ Rate of cooling during PQ
♦ is greater than
♦ Rate of cooling during UV
21. We can write this in two steps:
(i) In the initial stages, when the temperature is high,
♦ More heat is lost in any given duration
(ii) In the later stages, when the temper is low,
♦ Less heat is lost in that same duration
22. In other words:
• In the initial stages, when the 'temperature difference' is high,
♦ Rate of cooling is high
• In the later stages, when the 'temperature difference' is low,
♦ Rate of cooling is low
1. We see that, ‘rate of cooling’ is not a constant. It changes continuously
2. Sir Isaac Newton discovered that, the rate depends on the difference between two temperature values:
(i) The temperature of the body
♦ We can denote it as T
(ii) The temperature of the surroundings
♦ We can denote it as Ts
3. The rate is directly proportional to the difference (T - Ts)
• That means:
♦ If (T - Ts) is large, the rate will be large
A large (T - Ts) indicates a hot body
• So it is obvious that:
A body will lose more heat when it is ‘more hot’ than when it is ‘less hot’ (even when the time intervals are the same)
5. In addition to the above information, Newton discovered that:
• The rate of cooling depends on two more factors:
♦ The nature of surface of the body
♦ The area of the surface of the body
• The above two items are constants. So they can be represented by a single constant K
• Thus we get Eq.11.14:
Rate of cooling of a body = K (T - Ts)
6. Each body will have a unique value for K
• If we can find K, we will be able to calculate the rate of cooling
• If we get the rate of cooling, we will be able to calculate the time required for a body to cool down to any required temperature
• The following two solved examples will demonstrate such a procedure
Solved example 11.31
A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?
Solution:
Data given is:
♦ Temperatures at the initial stage: 94 °C and 86 °C
♦ Time for cooling = 2 minutes
♦ Temperatures at the final stage: 71 °C and 69 °C
♦ Time for cooling = ?
♦ Surrounding temperature, Ts = 20 °C
Case 1: The food cools from 94 °C to 86 °C
1. When the food is at 94 °C, it is hot
• When it is hot at 94 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{94-86}{2 \times 60}=\frac{4}{60}}}$
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{94+86}{2}=90 \right )}}$
3. So this rate occurs when the difference in temperature is (90 - 20) = 70 °C
4. Applying Eq.11.14, we get: $\mathbf\small{\rm{\frac{4}{60}}}$ = K × 70
• So K = $\mathbf\small{\rm{\frac{4}{60 \times 70}}}$
Case 2: The body cools from 71 °C to 69 °C
1. When the body is at 71 °C, it is less hot
• When it is less hot at 71 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{71-69}{t}=\frac{2}{t}}}$
♦ Where 't' is the time required to cool from 71 °C to 69 °C
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{71+69}{2}=70 \right )}}$
3. So this rate occurs when the difference in temperature is (70 - 20) = 50 °C
4. Applying Eq.11.14, we get: $\mathbf\small{\rm{\frac{2}{t}}}$ = K × 50
• But from case 1, we have the value of K
• Substituting that value of K, we get:
$\mathbf\small{\rm{\frac{2}{t}=\frac{4}{60 \times 70} \times 50}}$
• So t = 42 s
Solved example 11.32
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool
from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Solution:
Data given is:
♦ Temperatures at the initial stage: 80 °C and 50 °C
♦ Time for cooling = 5 minutes
♦ Temperatures at the final stage: 60 °C and 30 °C
♦ Time for cooling = ?
♦ Surrounding temperature, Ts = 20 °C
Case 1: The body cools from 80 °C to 50 °C
1. When the body is at 80 °C, it is hot
• When it is hot at 80 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{80-50}{5 \times 60}}}$ = 0.1
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{80+50}{2}=65 \right )}}$
3. So this rate occurs when the difference in temperature is (65 - 20) = 45 °C
4. Applying Eq.11.14, we get; 0.1 = K × 45
• So K = $\mathbf\small{\rm{\frac{0.1}{45}}}$
Case 2: The body cools from 60 °C to 30 °C
1. When the body is at 60 °C, it is less hot
• When it is less hot at 60 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{60-30}{t}=\frac{30}{t}}}$
♦ Where 't' is the time required to cool from 60 °C to 30 °C
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{60+30}{2}=45 \right )}}$
3. So this rate occurs when the difference in temperature is (45 - 20) = 25 °C
4. Applying Eq.11.14, we get; $\mathbf\small{\rm{\frac{30}{t}}}$ = K × 25
• But from case 1, we have the value of K
• Substituting that value of K, we get:
$\mathbf\small{\rm{\frac{30}{t}=\frac{0.1}{45} \times 25}}$
• So t = 540 s = 9 min
• When the food in solved example 11.31 is at 94 °C, the ‘rate of cooling’ will have a particular value
• When the temperature falls to 93 °C, the rate of cooling will be having another value
• In this way, the rate continuously changes
• We took the average temperature to find K
• But it is possible to find instantaneous rates by applying calculus
• Such a method will give more accurate results. We will see it after completing the basic course in calculus
PREVIOUS
CONTENTS
NEXT
Copyright©2020 Higher Secondary Physics. blogspot.in - All Rights Reserved
No comments:
Post a Comment