Newton's Law Of Cooling

Newton's Law of Cooling

Newton's law of cooling states that the rate at which an exposed body changes temperature through radiation is approximately proportional to the difference between the object's temperature and that of its surroundings, under the assumption of small differences.

Newton’s law of cooling is given by, dT/dt = k(T_t – T_s)

Where,

• T_t = Temperature of the body at time t

• T_s = Temperature of the surrounding
• k = Positive constant that depends on the area and nature of the surface of the body under consideration.

Newton’s Law of Cooling Formula

T(t) = T_s + (T_o – T_s) e^-kt

Where,

• t = time,
• T(t) = Temperature of the given body at time t
• T_s = Surrounding temperature
• T_o = Initial temperature of the body
• k = Constant

According to Newton's Law of Cooling

$\frac{d \theta}{d t} \alpha\left(\theta-\theta_0\right)$

$\text { or we can say that } \frac{d \theta}{d t}=k\left(\theta-\theta_0\right)$

where,

k is the proportionality constant

$\begin{aligned}
& R=\frac{d \theta}{d t}=\text { Rate of cooling } \\
& \theta=\text { Temperature of the body } \\
& \theta_0=\text { Temperature of the surrounding }
\end{aligned}$

Using the above formula we can plot various curves

1. The curve between $\log \left(\theta-\theta_0\right) V s \quad \text { time }(t)$

$\text { As } \log _e\left(\theta-\theta_0\right)=-k t+c$

So the graph will be

2. The curve between the Temperature of the body and time $\text { i.e } \theta Vs \quad t$

$\mathrm{As} \theta-\theta_0=A e^{-k_t}$

So the graph will be

3. The curve between the rate of Cooling and body temperature $I.e R=\frac{d \theta}{d t} \text { vs } \theta$

As,

$ R=\frac{d \theta}{d t}=K\left(\theta-\theta_0\right)=K \theta-K \theta_0$

So the graph will be

4. The curve between the Rate of Cooling (R) and the Temperature difference between body and Surrounding

$\text { I.e } R=\frac{d \theta}{d t} \quad V s \quad\left(\theta-\theta_0\right)$

$\text { As } R \alpha\left(\theta-\theta_0\right)$

So the graph will be

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Solved Example Based on Newton’s Law of Cooling

Example 1: If R represents the rate of heat radiation then the rate of heat transferred by two bodies at a temperature of 27^oC and 327^oC and their radii are in the ratio of 2:1 is

1) 4:1

2) 1:4

3) 16:1

4) 1:16

Solution:

if two bodies of the same material under identical environments -

$\begin{aligned}
& \frac{\left(R_c\right)_1}{\left(R_c\right)_2}=\frac{A_1 v_2}{A_2 v_1} \\
& \frac{R_1}{R_2}=\frac{A_1}{T_1^4} A_2 T_2^4=\left(\frac{r_1}{r_2}\right)^2 \cdot\left(\frac{T_1}{T_2}\right)^4=\left(\frac{2}{1}\right)^2 \cdot\left(\frac{300}{600}\right)^4=\frac{1}{4}
\end{aligned}$

Hence, the answer is option (2).

Example 2: Two bodies of the same material but different radii are heated up to the same temperature. $r_1>r_2$, where $r_1$ and $r_2$ are the radii of the two bodies then

1) Both will cool at the same rate

2) 1st will cool at a faster rate

3) 2nd will cool at a faster rate

4) None of the above

Solution:

We know,

Rate of cooling $\propto r^2$

$\therefore$ 1st will cool at a faster rate.

Hence, the answer is option (2).

Example 3: The energy emitted per second by a black body at 27^oC is 10J. If the temperature of the blackbody is increased to 327^oC, the energy (in Joule) emitted per second will be:

1) 20

2) 40

3) 80

4) 160

Solution

$ \begin{aligned}
& E \propto T^4 \\
& \frac{E_1}{E_2}=\left(\frac{T_2}{T_1}\right)^4 \Rightarrow E_2=E_1\left(\frac{T_2}{T_1}\right)^4 \\
& E_2=10 \times\left(\frac{273+327}{273+27}\right)^4=160 \mathrm{~J}
\end{aligned}$

Hence, the answer is option (4).

Example 4: A bucket full of water cools from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in time $T_1$, from $70^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in time $T_2$, and from $65^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in time $T_3$, then

1) $T_1=T_2=T_3$
2) $T_1>T_2>T_3$
3) $T_1<T_2<T_3$
4) $T_1>T_2<T_3$

Solution

Newton's Law of Cooling

The rate of cooling is directly proportional to the temperature difference between the body and its surroundings.

$\text { And When the body Cools by Radiation from } \theta_1^0 C_{\text {to theta }} \theta_2^0 C_{\text {in time t }}$

$\text { Then }\left[\frac{\theta_1-\theta_2}{t}\right]=k\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right] $

$
\begin{aligned}
& \because \text { Rate of cooling }=\left[\frac{d \theta}{t}\right] \propto\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right] \\
& \left(\frac{\theta_1+\theta_2}{2}-\theta_o\right)_1>\left(\frac{\theta_1+\theta_2}{2}-\theta_o\right)_2>\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right)_3 \\
& \text { So }\left(\frac{d \theta}{T_1}\right)_1>\left(\frac{d \theta}{T_2}\right)_2>\left(\frac{d \theta}{T_3}\right)_3
\end{aligned}
$

Hence $T_1<T_2<T_3$

Hence, the answer is option (3).

Example 5: Two identical beakers $\mathrm{A}$ and $\mathrm{B}$ contain equal volumes of two different liquids at $60^{\circ} \mathrm{C}$ each and left to cool down. Liquid in $\mathrm{A}$ has liquid in $\mathrm{B}$ has a density of $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and specific heat of $4000 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

1)

2)

3)

4)

Solution

Newton's Law of Cooling -

$\frac{d \theta}{d t} \alpha\left(\theta-\theta_0\right)$

- wherein

The temperature difference is not very large.

From given

$\begin{aligned}
&\begin{aligned}
& \rho_A<\rho_B, \\
& \Rightarrow m_A<m_B
\end{aligned}\\
&\text { and } \Rightarrow s_A<s_B \text {, }
\end{aligned}$

By newtons law of cooling

$
\begin{aligned}
& \frac{-d T}{d t}=\frac{4 \sigma e A T_0^3\left(T-T_o\right)}{m s} \\
& \Rightarrow \frac{-d T}{d t} \alpha \frac{1}{m s}
\end{aligned}
$
at $t=0$
$
\begin{aligned}
& -\left(\frac{d T}{d t}\right)_A \alpha \frac{1}{m_A s_A} \\
& -\left(\frac{d T}{d t}\right)_B \alpha \frac{1}{m_B s_B}
\end{aligned}
$

and we know

$m_A s_A<m_B s_B$

So the slope of the T v/s t curve for A is more than B.

Hence, the answer is option (3).

Summary

Newton's Law of Cooling explains how the temperature of an object decreases over time as it approaches the ambient temperature. The cooling rate is proportional to the temperature difference between the object and its surroundings. This principle applies to various real-life situations and is used in different fields like metallurgy, food science, and medicine. The law is mathematically represented and can be applied to solve problems involving heat transfer, rate of cooling, and temperature changes in objects over time.