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Temperature of a hot object tends to cool down faster as soon as it is detached from the heat source but after a certain time, it remains at constant temperature for a long duration. The object does not cool down beyond room temperature. This can be explained by observing the relationship between the temperature of a hot body and time by Newton’s Law of Cooling. The complete explanation about the cooling curve is done based on the experiment of Newton’s Law of Cooling.
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Key Takeaways- Newton’s Law of Cooling, Temperature, Cooling Curve, energy, heat, time, specific heat capacity.
Also Read: Heat Capacity
Newton’s Law of Cooling
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Newton’s law emphasizes the rate of cooling of the hot object which becomes the relationship between the temperature of the hot object and time. It states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings.
Newton’s Law of Cooling
Hot objects have high energy within them thus, they radiate quickly to their surroundings. Energy flows from the higher potential to the lower potential and as they keep on losing their energy, they start to cool down slowly. As the energy difference reduces, the rate of heat loss also reduces. The statement for Newton’s Law of Cooling can be written form as:
– \(\frac{dQ}{dt}\) = k (T2 – T1) … (1)
where,
Q is heat.
- \(\frac{dQ}{dt}\) denotes the rate of loss of heat.
t is time.
T1 is the temperature of the surrounding of the hot body.
T2 is the temperature of the hot body at that particular time.
k is a positive constant depending upon the nature and area of the surface of the body.
Also Read: Heat
Derivation of Newton’s Law of Cooling
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Let’s consider,
m – mass of the body
s – specific heat capacity
T1 – temperature of the surrounding of the hot body
T2 – temperature of the hot body at that particular time
dQ – amount of heat loss
dT2 – difference in temperature of the hot body as time changes
When the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,
dQ = ms dT2
The rate of loss of heat is given by,
\(\frac{dQ}{dt}\) = ms (\(\frac{dT^2}{dt}\)) … (2)
Comparing (1) and (2) –
-ms (\(\frac{dT^2}{dt}\)) = k (T2 – T1)
Rearranging the above equation
\(\frac {dT^2}{(T2-T1)}\)= – (\(\frac{k}{ms}\)) dt
\(\frac {dT^2}{(T2-T1)}\) = – Kdt
Where
K = \(\frac{k}{ms}\)
Integrating the above expression –
Loge (T2 – T1) = - Kt + c … (3)
Or
T2 = T1 + C`e-Kt
Where C` = ec
The above expression is used to calculate the time of cooling of a body through a particular range of temperature. The cooling curve is a graph that shows the relationship between body temperature and time. The rate of temperature fall is determined by the slope of the tangent to the curve at any point. Thus the slope includes the final temperature of the hot object (TH).
In simpler terms,
Tt = TA + (TH - TA) e-kt
Where
Tt is the temperature at time t,
TA is the ambient temperature (room temperature),
TH is the temperature of the hot object,
k is the positive constant and t is the time.
Also Read: Difference between Heat and Temperature
Experiment for Verifying Newton’s Law of Cooling
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Aim
To understand the relationship between the temperature of the hot body and time by plotting the cooling curve.
Experimental Setup
Apparatus
We will require the Newton’s law of cooling apparatus which includes
- Copper calorimeter
- Wooden lid with two holes, one for inserting thermometer and the other for stirrer, and
- Double-walled vessel
- Stop clock
- Hot water
- Cotton threads
- Beakers.
Procedure
- As shown in the figure, fill normal temperature water in the space between the double-walled enclosure. Place a thermometer to measure the ambient/ room temperature (TA)
- Make a note of the least count of the stopwatch and thermometer before starting the readings. Set the watch at zero.
- Fill the beaker with hot water that is heated more than 800C and cover it with a lid which already has a thermometer and a stirrer attached to it.
- Start your readings once the temperature reaches 800C. Take readings every 2 minutes. Let your time interval increase when you see the water taking time to reach the room temperature.
- Stop your readings when you see the water almost reach the room temperature.
Graph and Observations
Record your observation according to the following table.
You will get a T0C vs t min graph as this –
Cooling curve graph
As per equation (3) for finding the ‘K’ the following graph is plotted. The negative slope thus obtained is the value of ‘K’
Conclusion
The experiment and its calculations go exactly as per Newton’s law of cooling as it is observed that in the beginning the temperature falls very quickly but slows down as it reaches the temperature that its surroundings have, i.e. the difference in temperature decreases.
Also Read: Thermal Properties of Matter
Limitations of Newton’s Law of Cooling
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- The loss of heat from the body can happen through many ways but Newton’s law of cooling takes forward to verify its law by considering that loss of heat takes place only through radiation.
- It is important to have smaller temperature differences between the body and its surroundings
- The temperature of the surrounding of the hot body should remain constant which is not always true in reality.
Applications of Newton’s Law of Cooling
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- Warm food kept in a large plate cools down faster than the food kept in a bowl as the rate of cooling is directly proportional to volume. The more the surface area of the food exposed to the surrounding temperatures, the faster it will cool down.
Newton’s Law of Cooling Application
- It is helpful in determining and predicting how long a particular body will take to cool down.
- By looking at a dead body, its time of death can be found by considering the current body temperature and the possible body temperature at the time of death.
Things to Remember
- It is important to have a uniform temperature throughout the hot body.
- The difference in temperature of the hot body at time intervals is taken to avoid any errors of the extremes.
- Newton’s law of cooling is reasonably accurate.
- The bigger the temperature difference between the object and its surroundings, the faster will the object lose its heat.
Also Read: Thermal Conductivity
Sample Questions
Ques. What is Newton’s law of Cooling? (3 marks)
Ans. Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings. According to Newton’s law of cooling, the rate of loss of heat, that is – dQ/dt of the body is directly proportional to the difference of temperature ΔT = (T2–T1) of the body and the surroundings. The law is only valid for small temperature differences. Furthermore, the amount of heat lost through radiation is determined by the composition of the body’s surface and the extent of the exposed surface.
Therefore, the expression can be written as:
– dQ/dt = k(T2–T1)
Ques. 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? (4 marks)
Ans. The average temperature of 94 °C and 86 °C is 90 °C, which is 70 °C above the room temperature. Under these conditions, the pan cools 8 °C in 2 minutes.
According to Newton’s law of cooling,
– dQ/dt = k(T2–T1)
Substitute the value in the above expression,
8 °C /2 min = k(70°C)…… (1)
The average of 69 °C and 71 °C is 70 °C, which is 50 °C above room temperature. the value of K is the same.
Substitute the value in the above expression,
2 °C /dt = k(50°C) … (2)
Equate equation (1) and (2),
dt = 0.7 min
or time is equal to 42 s.
Ques. Define specific heat capacity. (2 marks)
Ans. Specific heat capacity of a substance can be defined as the amount of heat required to raise the temperature of one mole of a substance through 10C. The amount of heat (q) of the substance is the product of the mass (m), specific heat (c), and difference in temperature of the body (ΔT). The specific heat capacity of water is 1 cal g-1 K-1
Ques. Why hot milk is easier to drink from a bowl than from a glass. (2 marks)
Ans. Bowl has a greater surface area than glass therefore more heat is lost to its surroundings in the form of heat radiation through the bowl. The cooling of hot water depends upon the difference between its temperature and the surroundings. The rate of cooling is faster at first and then slows as the temperature drops.
Ques. A body at temperature 40ºC is kept in a surrounding of constant temperature 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC. (4 marks)
Ans. According to Newton's law of cooling
Tt - TA = (TH - TA) e-kt
Now, for the interval in which temperature falls from 40 ºC to 35 ºC.
(35 – 20) = (40 – 20) e-(10k)
e-(10k) = ¾
-10k = (ln 4/3)
k = 0.2876/10
k = 0.02876
Now, for the next interval,
(30 – 20) = (35 – 20) e-kt
10 = 15e-kt
e-kt = 2/3
-kt = ln (2/3)
t = 0.40546/k
Substitute the value of k in the above equation,
t = 0.40546/0.02876
t = 14.098 min.
Ques. The oil is heated to 70 ºC. It cools to 50 ºC after 6 minutes. Calculate the time taken by the oil to cool from 50 ºC to 40 ºC given the surrounding temperature Ts = 25 ºC (4 marks)
Ans. Given,
The temperature of oil after 6 min i.e. T(t) is equal to 50 ºC.
The ambient temperature T is 25 ºC.
The temperature of oil, To is 70 ºC.
The time to cool to 50ºC is 6 min.
According to Newton’s law of cooling,
T(t) = T + (T0 – T) e-kt
(T(t) – T) / (To – T) = e-kt
-kt = ln[(T(t) – T) / (To – T)] ……… (1)
Substitute the above data in Newton’s law of cooling expression,
-kt = ln [(50 – 25) / (70 – 25)]
-k = (ln 0.55556)/6
k = 0.09796
The average temperature is equal to 45 ºC
Substitute the values in equation (1),
-(0.09796) t = ln [(45 – 25) / (70 – 25)]
-0.09796t = ln (0.44444)
0.09796t = 0.81093
t = 0.09796/0.58778
t = 8.278 min.
Ques. Water is heated to 80 ºC for 10 min. How much would the temperature be in degrees Celsius, if k = 0.056 per min and the surrounding temperature is 25 ºC? (4 marks)
Ans. Given,
The ambient temperature T is 25 ºC,
The temperature of water T0 is 80 ºC.
The time that water is heated is t 10 min.
The value of constant k is 0.056.
According to Newton’s law of cooling,
T(t) = T + (T0 – T) e-kt
Substitute the above data in the above expression,
T(t)= 25 + (80 – 25) e-(0.056×10)
T(t) = 25+55 e-(0.056×10)
T(t) = 25+31.42
T(t) = 56.42
After 10 min the temperature cools down from 80 ºC to 56.42 ºC.
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