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Thermal Physics

Newton's Law of Cooling

Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings. It is an approximate consequence of Stefan-Boltzmann law applicable for small temperature excesses.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Direct numerical calculations involving cooling time, average temperature method, and temperature versus time graphs.

PGT extension

Differential equation formulation, exponential temperature decay, and connection with Stefan-Boltzmann law of radiation.

Prerequisites

Concept of Temperature
Modes of heat transfer: Conduction, Convection, and Radiation
Specific heat capacity and thermal capacity

Physical Basis of Newton's Law of Cooling

The law is derived from Stefan-Boltzmann law of radiation under the condition that the temperature difference between the body and surroundings is small.
It assumes that the loss of heat is solely due to radiation and that forced convection conditions or surrounding temperature remain strictly constant.
The rate of cooling depends on the nature of the radiating surface (emissivity) and the total surface area exposed.

Mathematical Formulation

Let a body of mass m, specific heat s, and surface area A be at temperature T, while the surrounding medium is at constant temperature T_0.
The rate of heat loss dQ/dt is proportional to (T - T_0), leading to dQ/d$t = -K(T - T_0)$, where K is a positive constant.
Since dQ = ms dT, the rate of fall of temperature is given by dT/d$t = -k(T - T_0)$, where $k = K/(ms)$.

Average Temperature Method in Numerical Problems

For moderate temperature differences, the rate of cooling dT/dt can be approximated as the change in temperature divided by time interval: (T_1 - $T_2)/t = k(bar{T} - T_0)$.
Here, bar{T} = (T_1 + T_2)/2 represents the mean temperature of the body during the time interval t.
This approximation is extensively used in UP TGT/PGT examinations to bypass complex exponential integration.

Exponential Decay of Temperature

Integrating the differential equation dT/d$t = -k(T - T_0) yields T(t) = T_0 + (T_i - T_0)e^{-kt}$, showing an exponential approach to ambient temperature.
The temperature of the body theoretically reaches the ambient temperature only as time approaches infinity.
The time constant of the cooling process is ta$u = 1/k$, which signifies the time required for the excess temperature to reduce to 1/e of its initial value.

Derivations

Derivation of Exponential Temperature Decay

Conditions: Small temperature difference between body and surroundings, constant ambient temperature.

Write the fundamental differential equation for temperature decay: \frac{dT}{dt} = -k(T - T_0).
Separate the variables T and t: \frac{dT}{T - $T_0 = -k dt$.
Integrate both sides with initial condition $T = T_i at t = 0: \int_{T_i}^{T} \frac{dT}{T - T_0} = -k \int_{0}^{t} dt$.
Evaluate the integrals: \ln\left(\frac{T - T_0}{T_i - T_0}\right) = -kt.
Exponentiate both sides to obtain the final explicit temperature function.

Result: T(t) = T_0 + (T_i - T_0)e^{-kt}

Approximation of Newton's Law from Stefan-Boltzmann Law

Conditions: Body temperature T is slightly greater than surrounding temperature T_0 (i.e., T - T_0 \ll T_0).

According to Stefan-Boltzmann law, net rate of heat loss is \frac{dQ}{dt} = \sigma A e (T^4 - T_0^4).
Factor the temperature term: T^4 - $T_0^4 = (T^2 - T_0^2)(T^2 + T_0^2) = (T - T_0)(T + T_0)(T^2 + T_0^2)$.
Substitute T \approx T_0 into the binomial sum terms: T^4 - T_0^4 \approx (T - $T_0)(2T_0)(2T_0^2) = 4T_0^3(T - T_0)$.
Substitute back into the radiation equation: \frac{dQ}{dt} = 4 \sigma A e T_0^3 (T - T_0).
Since dQ = ms dT, the cooling rate becomes proportional to (T - T_0) with proportionality constant $K = 4 \sigma A e T_0^3 / (ms)$.

Result: \frac{dT}{dt} = -\frac{4 \sigma A e T_0^3}{ms} (T - T_0)

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Differential Law of Cooling	$$\frac{dT}{dt} = -k(T - T_0)$$	T = temperature of body, $T_0 = temperature of surroundings$, k = constant K s^{-1}	Small temperature difference
Average Temperature Form	$$\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$$	T_1, $T_2 = initial$ and final temperatures in time t, $T_0 = surrounding temperature$ K s^{-1}	Valid for small time intervals
Exponential Temperature Equation	$$T(t) = T_0 + (T_i - T_0)e^{-kt}$$	$T_i = initial temperature at t = 0$, t = time K	Exact solution of linear cooling differential equation
Cooling Constant	$$k = \frac{4\sigma A e T_0^3}{ms}$$	\sigma = Stefan's constant, A = surface area, e = emissivity, m = mass, s = specific heat s^{-1}	Derived from Stefan-Boltzmann law
Time Constant	$$\tau = \frac{1}{k}$$	\tau = time constant s	Time taken for excess temperature to decay to 1/e of initial value
Heat Loss Rate	$$\frac{dQ}{dt} = -ms k (T - T_0)$$	Q = heat energy, m = mass, s = specific heat J s^{-1} \text{ or W}	Equivalent to rate of fall of temperature multiplied by thermal capacity
Logarithmic Form of Temperature Decay	$$\ln\left(\frac{T - T_0}{T_i - T_0}\right) = -kt$$	Natural logarithm of normalized temperature excess Dimensionless	Rearranged exponential form
Ratio of Cooling Rates for Two Bodies	$$\frac{(dT/dt)_1}{(dT/dt)_2} = \frac{k_1 (T_1 - T_0)}{k_2 (T_2 - T_0)}$$	Subscripts 1 and 2 denote different bodies Dimensionless	Comparing cooling rates at given temperatures

Graphs and experiments

Temperature versus Time Curve

Draw: An exponential decay curve starting at T_i and asymptotically approaching the horizontal asymptote $T = T_0$ as t increases.

Interpret: The slope of the tangent at any point represents the rate of cooling dT/dt at that specific instant, which continuously decreases as the body cools.

Exam use: Frequently asked in identifying correct qualitative shapes of cooling curves versus heating curves.

Logarithm of Temperature Excess versus Time

Draw: A straight line with a negative slope on a semi-logarithmic plot of \ln(T - T_0) versus t.

Interpret: The slope of the straight line equals -k, confirming the exponential nature of Newton's law of cooling.

Exam use: Used to determine the experimental cooling constant k from laboratory data.

Rate of Cooling versus Temperature Difference

Draw: A straight line passing through the origin on a plot of dT/dt against (T - T_0).

Interpret: The direct proportionality between the rate of cooling and the temperature excess is verified by the constant linear slope k.

Exam use: Tested in conceptual questions to check validity limits of linear cooling.

Solved numericals

TGT core

A body cools from 80 \textdegree C to 50 \textdegree C in 5 minutes. Calculate the time it takes to cool from 60 \textdegree C to 30 \textdegree C, if the room temperature is 20 \textdegree C.

Use the average temperature formula: \frac{T_1 - T_2}{t} = k \left(\frac{T_1 + T_2}{2} - T_0\right).
For the first case: \frac{80 - 50}{5} = k \left(\frac{80 + 50}{2} - 20\right) \implies \frac{30}{5} = k (65 - 20) \implies $6 = k(45) \implies k = \frac{6}{45} = \frac{2}{15} \text{ min}^{-1}$.
For the second case, cooling from 60 \textdegree C to 30 \textdegree C in time t: \frac{60 - 30}{t} = k \left(\frac{60 + 30}{2} - 20\right) \implies \frac{30}{t} = \frac{2}{15}(45 - 20) \implies \frac{30}{t} = \frac{2}{15}(25) = \frac{10}{3}.
Solve for t: t = \frac{30 \times 3}{10} = 9 \text{ minutes}.

Answer: 9 minutes

PGT extension

A liquid cools from 70 \textdegree C to 60 \textdegree C in 5 minutes. The surrounding temperature is 30 \textdegree C. Find the temperature of the liquid after the next 5 minutes.

Apply the average temperature formula for the first interval (70 to 60 \textdegree C): \frac{70 - 60}{5} = k \left(\frac{70 + 60}{2} - 30\right) \implies $2 = k(65 - 30) \implies 2 = k(35) \implies k = \frac{2}{35} \text{ min}^{-1}$.
Let the temperature after the next 5 minutes be T. For the second interval (60 to T \textdegree C in 5 minutes): \frac{60 - T}{5} = k \left(\frac{60 + T}{2} - 30\right).
Substitute k: \frac{60 - T}{5} = \frac{2}{35} \left(\frac{60 + T - 60}{2}\right) = \frac{2}{35} \left(\frac{T}{2}\right) = \frac{T}{35}.
Simplify and solve for T: 7(60 - T) = T \implies 420 - 7T = T \implies 8T = 420 \implies T = \frac{420}{8} = 52.5 \text{ \textdegree C}.

Answer: 52.5 \textdegree C

TGT core

A metal sphere cools from 60 \textdegree C to 50 \textdegree C in 10 minutes. If the room temperature is 25 \textdegree C, what will be its temperature after another 10 minutes?

First interval: average temperatur$e = (60 + 50)/2 = 55 \textdegree C$. Temperature exces$s = 55 - 25 = 30 \textdegree C$. Rate of fal$l = (60 - 50)/10 = 1 \textdegree C/min$. Therefore, $1 = k(30) \implies k = 1/30 \text{ min}^{-1}$.
Second interval: let final temperature be T after another 10 minutes (cooling from 50 \textdegree C to T \textdegree C).
Average temperatur$e = (50 + T)/2$. Excess temperatur$e = (50 + T)/2 - 25 = T/2$.
Rate of fal$l = (50 - T)/10 = k (T/2) = \left(\frac{1}{30}\right) \left(\frac{T}{2}\right) = \frac{T}{60}$.
Cross multiply: 6(50 - T) = T \implies 300 - 6T = T \implies 7T = 300 \implies T = \frac{300}{7} \approx 42.86 \text{ \textdegree C}.

Answer: 42.86 \text{ \textdegree C}

Assumptions and limitations

The temperature difference between the body and surroundings must be small (typically less than 20 \textdegree C to 30 \textdegree C) for the linear approximation to hold accurately.
Heat loss is assumed to occur purely by radiation under uniform convective conditions; forced drafts or varying ambient temperatures invalidate the simple constant k assumption.

Common misconceptions

Common error: Newton's law of cooling applies for any arbitrary large temperature difference.

Correct: It is valid only for small temperature differences; for large differences, the full Stefan-Boltzmann radiation law must be used.

Linearization of T^4 - T_0^4 is valid via binomial expansion only when (T - T_0) \ll T_0.

Common error: Equal drops in temperature take equal intervals of time.

Correct: Equal drops in temperature take increasingly longer time intervals as the body approaches room temperature.

The rate of cooling is proportional to the temperature excess, which decreases exponentially with time.

Common error: A body cools at a constant rate throughout its cooling cycle.

Correct: The cooling rate decreases continuously as the body's temperature drops towards the ambient temperature.

The driving force for cooling is the temperature difference (T - T_0), which decreases monotonically.

Quick revision

Newton's law of cooling states that dT/d$t = -k(T - T_0)$, implying exponential temperature decay towards the ambient temperature T_0.
Differential Law of Cooling: \frac{dT}{dt} = -k(T - T_0)
Average Temperature Form: \frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)

Must remember

Newton's Law of Cooling is a special linear approximation of the Stefan-Boltzmann law valid only for small temperature differences between the body and its surroundings.
The average temperature method $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ is extensively used in UP TGT/PGT examinations to bypass complex exponential integration.
The time constant $\tau = 1/k$ signifies the time required for the excess temperature of a cooling body to reduce to $1/e$ of its initial value.
Equal drops in temperature take increasingly longer time intervals because the cooling rate decreases monotonically as the body approaches ambient temperature.

Glossary

$T_0$Constant temperature of the surrounding medium or ambient temperature.

$k$Cooling constant dependent on surface area, mass, specific heat, emissivity, and ambient temperature ($s^{-1}$).

$\tau$Time constant of the cooling process, defined as $\tau = 1/k$.

$\bar{T}$Mean or average temperature of the body during a given time interval, $(T_1 + T_2)/2$.

Newton's Law of Cooling vs. Stefan-Boltzmann Law

Parameter	Newton's Law of Cooling	Stefan-Boltzmann Law
Temperature Difference	Small temperature differences ($(T - T_0) \ll T_0$)	Any arbitrary or large temperature difference
Mathematical Form	$\frac{dT}{dt} = -k(T - T_0)$	$\frac{dQ}{dt} = \sigma A e (T^4 - T_0^4)$
Nature of Equation	Linear differential equation	Non-linear fourth-power equation

Practice Quiz

Choose an option to reveal the explanation.

1. Newton's Law of Cooling is derived from which fundamental law under the condition of small temperature differences?

Explanation: Newton's law of cooling is derived directly from the Stefan-Boltzmann law by linearizing the $T^4 - T_0^4$ term using binomial approximation when the temperature difference is small.

2. What is the physical dimension of the cooling constant $k$ in the equation $\frac{dT}{dt} = -k(T - T_0)$?

Explanation: Since $\frac{dT}{dt}$ has units of $K s^{-1}$ and $(T - T_0)$ has units of $K$, the constant $k$ must have units of $s^{-1}$.

3. In the average temperature approximation method, the mean temperature $\bar{T}$ during time interval $t$ when temperature falls from $T_1$ to $T_2$ is given by:

Explanation: The average temperature method assumes linear variation over small intervals, taking the arithmetic mean $\bar{T} = \frac{T_1 + T_2}{2}$.

4. The time constant $\tau$ of a cooling body is related to the cooling constant $k$ as:

Explanation: The time constant $\tau$ is defined as the reciprocal of the cooling constant, i.e., $\tau = \frac{1}{k}$, representing the time required for the excess temperature to reduce to $1/e$ of its initial value.

5. If a body cools from $80^{\circ}\text{C}$ to $60^{\circ}\text{C}$ in $5\text{ minutes}$ in a room at $20^{\circ}\text{C}$, what is the average temperature $\bar{T}$ used in the cooling equation?

Explanation: The average temperature is $\bar{T} = \frac{80 + 60}{2} = 70^{\circ}\text{C}$.

6. Which of the following represents the exact exponential decay of temperature according to Newton's law?

Explanation: Integrating $\frac{dT}{dt} = -k(T - T_0)$ yields $T(t) = T_0 + (T_i - T_0)e^{-kt}$, which approaches $T_0$ as $t \to \infty$.

7. Why do equal drops in temperature take increasingly longer time intervals as a body cools down?

Explanation: The rate of cooling $\frac{dT}{dt}$ is directly proportional to the temperature excess $(T - T_0)$. As the body cools, this excess decreases, slowing down the rate of temperature drop.

8. The cooling constant $k$ is given by $k = \frac{4\sigma A e T_0^3}{ms}$. This expression shows that $k$ depends directly on:

Explanation: In the expression $k = \frac{4\sigma A e T_0^3}{ms}$, $k$ is directly proportional to surface area $A$, emissivity $e$, and inversely proportional to mass $m$ and specific heat $s$.

9. A body is cooling in a room. The rate of fall of temperature $\frac{dT}{dt}$ at any instant is proportional to:

Explanation: By definition, Newton's law of cooling states that the rate of fall of temperature is proportional to the temperature difference between the body and its surroundings.

10. What is the theoretical time taken by a body to reach the exact ambient temperature $T_0$?

Explanation: Since the temperature decay is exponential ($e^{-kt}$), the body reaches ambient temperature theoretically only as $t \to \infty$.

11. If the temperature difference between a body and its surroundings is doubled, the rate of cooling under Newton's law will:

Explanation: Since $\frac{dT}{dt} = -k(T - T_0)$, the rate of cooling is linearly proportional to the temperature difference $(T - T_0)$. Doubling the difference doubles the rate of cooling.

12. Which assumption is NOT made in standard Newton's law of cooling?

Explanation: The law assumes the nature of the radiating surface (emissivity) remains constant, not varying continuously with time.

13. The logarithmic form of normalized temperature excess in Newton's law of cooling is given by:

Explanation: Rearranging the exponential equation $T(t) = T_0 + (T_i - T_0)e^{-kt}$ yields $\frac{T - T_0}{T_i - T_0} = e^{-kt}$, taking natural log gives $\ln\left(\frac{T - T_0}{T_i - T_0}\right) = -kt$.

14. If two different bodies of identical surface area and mass are placed in identical surroundings, the one with higher specific heat $s$ will cool:

Explanation: Since $k = \frac{4\sigma A e T_0^3}{ms}$, a larger specific heat $s$ results in a smaller cooling constant $k$, meaning the body cools slower.

15. Which graph best represents the variation of temperature $T$ of a body with time $t$ during cooling?

Explanation: The temperature-time graph is an exponential decay curve that asymptotically approaches the ambient temperature $T_0$.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. A liquid cools from $70^{\circ}\text{C}$ to $60^{\circ}\text{C}$ in $5\text{ minutes}$ in a surrounding of temperature $30^{\circ}\text{C}$. What will be its temperature after the next $5\text{ minutes}$?

Explanation: Using the average temperature method: $\frac{70 - 60}{5} = k\left(\frac{70 + 60}{2} - 30\right) \implies 2 = k(65 - 30) = 35k \implies k = 2/35$. For the next interval from $60^{\circ}\text{C}$ to $T_2$ in $5\text{ min}$: $\frac{60 - T_2}{5} = \frac{2}{35} \left(\frac{60 + T_2}{2} - 30\right) = \frac{2}{35}\left(\frac{T_2}{2}\right) = \frac{T_2}{35}$. Solving gives $7(60 - T_2) = T_2 \implies 420 = 8T_2 \implies T_2 = 52.5^{\circ}\text{C}$.

2. A body cools from $50^{\circ}\text{C}$ to $40^{\circ}\text{C}$ in $5\text{ minutes}$. The surrounding temperature is $20^{\circ}\text{C}$. In what further time will it cool down to $30^{\circ}\text{C}$?

Explanation: First interval: $\frac{50 - 40}{5} = k(\frac{50+40}{2} - 20) \implies 2 = k(45 - 20) = 25k \implies k = 2/25$. Second interval from $40$ to $30^{\circ}\text{C}$ in time $t$: $\frac{40 - 30}{t} = \frac{2}{25}(\frac{40+30}{2} - 20) \implies \frac{10}{t} = \frac{2}{25}(35 - 20) = \frac{2}{25}(15) = \frac{30}{25} = \frac{6}{5}$. Thus $t = \frac{50}{6} = \frac{25}{3}\text{ minutes}$.

3. The initial temperature excess of a body over its surroundings is $T_i - T_0$. If the time constant of the cooling body is $\tau$, what is the temperature excess after time $t = \tau$?

Explanation: The temperature excess at any time $t$ is given by $(T(t) - T_0) = (T_i - T_0)e^{-kt}$. At $t = \tau = 1/k$, the excess becomes $(T_i - T_0)e^{-1} = \frac{T_i - T_0}{e}$ accroding to the definition of the time constant.

4. Two identical metallic spheres of same material and surface finish are heated to $100^{\circ}\text{C}$ and allowed to cool in identical surroundings of $20^{\circ}\text{C}$. If one sphere is solid and the other is hollow of the same outer radius, which one cools faster initially?

Explanation: Since $k = \frac{4\sigma A e T_0^3}{ms}$, and the hollow sphere has a smaller mass $m$ for the same surface area $A$ and specific heat $s$, its cooling constant $k$ is larger. Thus, the hollow sphere cools faster.

5. A body takes $4\text{ minutes}$ to cool from $90^{\circ}\text{C}$ to $80^{\circ}\text{C}$. If the room temperature is $20^{\circ}\text{C}$, the time taken by the body to cool from $60^{\circ}\text{C}$ to $50^{\circ}\text{C}$ will be:

Explanation: The average temperature for the first interval is $\bar{T}_1 = 85^{\circ}\text{C}$ (excess = $65^{\circ}\text{C}$). For the second interval ($60^{\circ}\text{C}$ to $50^{\circ}\text{C}$), average temperature is $\bar{T}_2 = 55^{\circ}\text{C}$ (excess = $35^{\circ}\text{C}$). Since the temperature excess is smaller in the second case, the rate of cooling is smaller, so it takes more time.

6. If the rate of fall of temperature of a body is $0.5^{\circ}\text{C/min}$ when its temperature is $60^{\circ}\text{C}$ and surrounding temperature is $20^{\circ}\text{C}$, what is the value of the cooling constant $k$?

Explanation: Using $\frac{dT}{dt} = -k(T - T_0) \implies 0.5 = k(60 - 20) = 40k \implies k = \frac{0.5}{40} = \frac{1}{80} = 0.0125 \text{ min}^{-1}$ (Wait, check calculation: $0.5 / 40 = 0.0125$. Let's ensure correct option index: Option A is $0.0125$). Let's set correct_index to 0.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. Which of the following differential equations correctly represents Newton's Law of Cooling for temperature $T$ and ambient temperature $T_0$?

Explanation: Newton's law states that rate of fall of temperature is proportional to $(T - T_0)$, leading to $\frac{dT}{dt} = -k(T - T_0)$.

2. If the cooling constant $k$ of a body depends on mass $m$ and specific heat $s$, how does an increase in specific heat affect the cooling rate?

Explanation: A higher specific heat $s$ means greater thermal inertia, which reduces the cooling constant $k$ and thereby decreases the cooling rate.

3. A liquid cools from $60^{\circ}\text{C}$ to $50^{\circ}\text{C}$ in $10\text{ minutes}$ when the room temperature is $25^{\circ}\text{C}$. Find the time taken to cool from $50^{\circ}\text{C}$ to $40^{\circ}\text{C}$ under the same conditions.

Explanation: First interval: $\frac{60 - 50}{10} = k(55 - 25) \implies 1 = 30k \implies k = 1/30$. Second interval ($50$ to $40^{\circ}\text{C}$): $\frac{50 - 40}{t} = \frac{1}{30}(45 - 25) \implies \frac{10}{t} = \frac{20}{30} = \frac{2}{3} \implies t = 15 \times \text{wait, } 10/t = 2/3 \implies 2t = 30 \implies t = 15$ min? Let's check: $45 - 25 = 20$. $\frac{1}{30} \times 20 = \frac{2}{3}$. $\frac{10}{t} = \frac{2}{3} \implies t = 15$. Wait, let's recalculate accurately: Average temp for first: $55$, excess $= 30$. $10/10 = 1 = k(30) \implies k = 1/30$. Second average: $45$, excess $= 20$. $\frac{10}{t} = \frac{1}{30}(20) = \frac{2}{3} \implies t = 15\text{ min}$. Wait, let's adjust option C to 15 minutes.

4. What is the primary condition for the validity of Newton's Law of Cooling derived from Stefan-Boltzmann law?

Explanation: The binomial expansion of $(T^4 - T_0^4)$ is valid only when the temperature difference $(T - T_0)$ is much smaller than the ambient temperature $T_0$.

5. If the time constant of a given body is $200\text{ seconds}$, its cooling constant $k$ is:

Explanation: Since $k = 1/\tau$, we have $k = \frac{1}{200} = 0.005 \text{ s}^{-1}$.

6. When plotting $\ln(T - T_0)$ against time $t$ for a body obeying Newton's law of cooling, what type of curve is obtained?

Explanation: From $\ln(T - T_0) = -kt + \ln(T_i - T_0)$, the graph is a straight line with slope $-k$ and y-intercept $\ln(T_i - T_0)$.

7. If two bodies A and B have cooling constants $k_A$ and $k_B$ such that $k_A = 2 k_B$, which body loses temperature faster under identical temperature excesses?

Explanation: Since the rate of cooling is directly proportional to $k$ ($\frac{dT}{dt} = -k(T - T_0)$), a larger $k_A$ means Body A cools faster than Body B.

8. Which of the following parameters does NOT directly affect the cooling constant $k$ of a solid metallic sphere?

Explanation: The cooling constant $k = \frac{4\sigma A e T_0^3}{ms}$ depends on ambient temperature $T_0$, surface area, mass, specific heat, and emissivity, but not on the initial temperature $T_i$ from which it started cooling.

9. The heat loss per unit time $\frac{dQ}{dt}$ of a cooling body is related to the rate of fall of temperature $\frac{dT}{dt}$ by:

Explanation: Since $dQ = ms dT$, dividing by $dt$ gives the rate of heat loss $\frac{dQ}{dt} = ms \frac{dT}{dt}$.

10. In forced convection conditions, why does Newton's Law of Cooling fail or require modification?

Explanation: Newton's law assumes constant idealized conditions (primarily radiation/natural cooling). Forced convection significantly alters the heat transfer coefficient, making the simple constant $k$ inadequate.