Q.A pan filled with hot food cools from $94\ ^\circ\text{C}$ to $86\ ^\circ\text{C}$ in $2$ minutes when the room temperature is at $20\ ^\circ\text{C}$. How long will it take to cool from $71\ ^\circ\text{C}$ to $69\ ^\circ\text{C}$?

Concept and Intuition

Newton's law of cooling states the rate of loss of temperature is proportional to the excess of the body's temperature over its surroundings: $\dfrac{dT}{dt} = -k(T-T_s)$. For a finite time interval, this is commonly approximated using the average temperature of the body during that interval:

Applying this twice (over the first 10 minutes, then the next 10 minutes) gives two equations in the two unknowns $k$ and $T_s$; dividing eliminates $k$ and lets us solve directly for $T_s$.

Step-by-Step Solution

1. First 10 minutes: temperature falls $60^0C\to50^0C$. Average temperature $=55^0C$. Rate $= \dfrac{10}{10}=1^0C/\text{min}$.

2. Next 10 minutes: temperature falls $50^0C\to42^0C$. Average temperature $=46^0C$. Rate $=\dfrac{8}{10}=0.8^0C/\text{min}$.

3. Divide (i) by (ii): …

Concept and Intuition

Newton's law of cooling in its practical (finite-interval) form states that the average rate of temperature fall equals a constant $k$ times the difference between the average body temperature over that interval and the surrounding temperature: $\dfrac{T_1-T_2}{t} = k\left[\dfrac{T_1+T_2}{2}-T_s\right]$. This lets us use the given cooling data (80°C to 50°C in 5 min) to solve for $k$ (a property of the body/surroundings that stays fixed), then reuse that $k$ for the second cooling interval (60°C to 30°C) to find the unknown time.

Step-by-Step Solution

1. First interval: $T_1=80°C$, $T_2=50°C$, $t=5\ min$, $T_s=20°C$. $\dfrac{80-50}{5} = k\left[\dfrac{80+50}{2}-20\right] \Rightarrow 6 = k(65-20)=45k \Rightarrow k = \dfrac{6}{45}=\dfrac{2}{15}\ \mathrm{min^{-1}}$.
2. Second interval: $T_1=60°C$, $T_2=30°C$, $T_s=20°C$, unknown $t$. $\dfrac{60-30}{t} = k\left[\dfrac{60+30}{2}-20\right] = k(45-20)=25k$. …

Concept and Intuition

Newton's law of cooling states the rate of loss of heat (and hence of temperature) is proportional to the excess of the body's temperature over its surroundings: $-\dfrac{dT}{dt} = k(T-T_0)$. For a body cooling by a modest amount, engineers/exam-setters use the convenient average-temperature approximation:

where $T_1,T_2$ are the temperatures at the start and end of an interval of duration $t$, and $T_0$ is the (constant) surrounding temperature. Applying this to two consecutive intervals gives two equations with two unknowns, $k$ and $T_0$.

Step-by-Step Solution

1. First interval ($62°\mathrm{C}\to50°\mathrm{C}$ in 10 min):

2. Second interval ($50°\mathrm{C}\to42°\mathrm{C}$ in the next 10 min): $\frac{50-42}{10} = k\left(\frac{50+42}{2}-T_0\right) \Rightarrow 0.8 = k(46-T_0) \quad (ii)$ …

Concept and Intuition

Newton's law of cooling (rate $\propto T-T_0$) is only a linear approximation valid for small temperature differences. Here the body's temperature (600 K, 900 K) is far above the ambient (300 K), so the full Stefan–Boltzmann result applies: net radiative cooling rate $\propto T^4-T_0^4$.

Step-by-Step Solution

1. Rate of cooling at temperature $T$ with surroundings at $T_0$: $\text{Rate}\propto T^4-T_0^4$.
2. Write $T_0=300$ K as the unit; $600=2T_0$ and $900=3T_0$.
3. At $600$ K: $\text{Rate}_1\propto (2T_0)^4-T_0^4=T_0^4(16-1)=15T_0^4=H$.
4. At $900$ K: $\text{Rate}_2\propto (3T_0)^4-T_0^4=T_0^4(81-1)=80T_0^4$. …

Concept and intuition

Newton’s Law of Cooling says that a body’s temperature changes at a rate proportional to the difference between its own temperature and the surrounding temperature. If the body is hotter than the surroundings, it cools down; if colder, it warms up. The differential equation

captures this: the constant $k>0$ controls how fast the temperature approaches $T_s$. The solution must be an exponential approach, because the driving force (the temperature difference) shrinks as $T$ gets closer to $T_s$.

1. Set up the differential equation We have

This is a separable ODE. Rewrite it as

2. Integrate both sides

The left side integrates to $-\ln|T_s - T|$ (by the substitution $u = T_s - T$, $du = -dT$). So

where $C$ is the constant of integration.

3. Solve for $T$ Multiply both sides by $-1$:

Exponentiate:

Let $A = e^{-C}$ (a positive constant). Then

Since $T_s - T$ can be positive or negative depending on whether $T$ is below or above $T_s$, we absorb the sign into a new constant $B$:

where $B$ can be any real number.

4. Apply the initial condition …

Concept and Intuition

Newton's Law of Cooling states the rate of heat loss is proportional to the temperature difference between the body and its surroundings. For a finite time interval, a good approximation replaces the instantaneous temperature with the average temperature over that interval: $\dfrac{\Delta T}{\Delta t} = k(T_{avg} - T_{room})$.

Step-by-Step Solution

1. First interval: cools from 200°C to 100°C in 5 minutes. $\Delta T = 100$, average temperature $=\dfrac{200+100}{2}=150^\circ\mathrm{C}$.
2. Apply the law: $\dfrac{100}{5} = k(150-30) \Rightarrow 20 = 120k \Rightarrow k = \dfrac{1}{6}\ \mathrm{min^{-1}}$.
3. Second interval: cools from 100°C to 50°C, $\Delta T = 50$, average temperature $= \dfrac{100+50}{2}=75^\circ\mathrm{C}$. …

Concept and Intuition

Newton's law of cooling says $\dfrac{dT}{dt}\propto (T-T_{surroundings})$ — the hotter the object is

relative to its environment, the faster it loses heat. As the body's own temperature drops, that

excess $(T-T_{surroundings})$ shrinks, so cooling naturally slows down even though the size of the

temperature interval (10$^\circ$C each time) stays the same. This is exactly why a hot cup of tea

cools quickly at first and much more slowly as it nears room temperature.

Step-by-Step Solution

1. From 75$^\circ$C to 65$^\circ$C, the average temperature is around 70$^\circ$C — well above the (unstated but lower) room temperature, so the excess $(T-T_{surroundings})$ is relatively large, giving a fast rate and hence a 10-minute drop.
2. From 65$^\circ$C to 55$^\circ$C, the average temperature (~60$^\circ$C) is closer to room temperature, so the excess is smaller, the rate of cooling is slower. …

Concept and Intuition

The exact rate of heat loss from a body by radiation follows Stefan's law, $\dfrac{dQ}{dt}\propto (T^4 - T_s^4)$, which is nonlinear in temperature. When $T$ is only slightly greater than the surrounding temperature $T_s$, this expression can be expanded (via a binomial/Taylor approximation) and reduces to a much simpler linear form, $\dfrac{dQ}{dt}\propto (T-T_s)$ — this simplified linear relation is Newton's law of cooling. Because it relies on that approximation, it stops being accurate once the temperature difference becomes large.

Step-by-Step Solution

1. Newton's law of cooling states: rate of loss of heat (or temperature) $\propto (T - T_s)$, where $T$ is the body's temperature and $T_s$ is the surrounding temperature.
2. This linear form is derived by approximating the exact Stefan's-law radiative cooling expression $\propto(T^4-T_s^4)$ for the case $T \approx T_s$.
3. The approximation (binomial expansion, keeping only the first-order term) is valid only when $(T-T_s)$ is small compared to $T_s$ itself. …

Concept and Intuition

Newton's law of cooling states the cooling rate is proportional to the excess of the body's (average) temperature over the surroundings: $\dfrac{\theta_1-\theta_2}{t}=k\left(\dfrac{\theta_1+\theta_2}{2}-\theta_0\right)$.

Step-by-Step Solution

1. First interval ($52.5^\circ C\to47.5^\circ C$ in 5 min): $\dfrac{52.5-47.5}{5}=k\left(\dfrac{52.5+47.5}{2}-\theta_0\right)\Rightarrow 1=k(50-\theta_0)$.
2. Second interval ($47.5^\circ C\to42.5^\circ C$ in 7.5 min): $\dfrac{47.5-42.5}{7.5}=k\left(\dfrac{47.5+42.5}{2}-\theta_0\right)\Rightarrow \dfrac23=k(45-\theta_0)$.
3. Dividing the two equations: $\dfrac{1}{2/3}=\dfrac{50-\theta_0}{45-\theta_0}\Rightarrow\dfrac32=\dfrac{50-\theta_0}{45-\theta_0}$. …