NEET Newtons Law of Cooling MCQs & PYQs

According to Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the object and its surroundings. As the liquid cools, the temperature difference between the liquid and the surroundings decreases, which slows down the rate of cooling.

Therefore, it will take greater than 5 minutes to cool from 60°C to 50°C, as the temperature difference is smaller, resulting in a slower cooling rate.

Newton’s law of cooling is used in the laboratory to determine the specific heat of gases by measuring the rate of temperature change of a heated gas as it cools in a controlled environment. By observing the cooling curve and applying the law, the specific heat can be calculated based on the energy lost over time, as it depends on the heat capacity of the gas.

To derive the exponential equation for Newton's law of cooling, we start with the differential form of Newton's law:

\[
\frac{dT}{dt} = -k (T - T_s)
\]

where:
- \( T \) is the temperature of the object at time \( t \),
- \( T_s \) is the temperature of the surroundings (constant),
- \( k \) is a positive constant of proportionality.

Step 1: Separate Variables
We can rewrite the equation as:
\[
\frac{dT}{T - T_s} = -k \, dt
\]

 Step 2: Integrate Both Sides
Integrate both sides with respect to \( T \) and \( t \):
\[
\int \frac{1}{T - T_s} \, dT = -\int k \, dt
\]

This gives:
\[
\ln |T - T_s| = -kt + C
\]

where \( C \) is the integration constant.

Step 3: Exponentiate Both Sides
Exponentiate both sides to remove the logarithm:
\[
T - T_s = e^{-kt + C} = Ce^{-kt}
\]

where \( C = e^C \) is a new constant.

Step 4: Apply Initial Condition
Let \( T(0) = T_0 \), where \( T_0 \) is the initial temperature of the object. Then:
\[
T_0 - T_s = Ce^0 = C
\]

Thus, \( C = T_0 - T_s \), and the solution becomes:
\[
T = T_s + (T_0 - T_s)e^{-kt}
\]

Final Equation
\[
T(t) = T_s + (T_0 - T_s)e^{-kt}
\]

This is the exponential cooling equation that describes the temperature \( T \) of the object over time according to Newton's law of cooling. So the graph is exponentially decreasing function

According to Newton's law of cooling, \(\frac{T_1 - T_2}{\Delta t} = K\left(\frac{T_1+T_2}{2} - T_0\right)\). For the first interval, \(\frac{10}{t} = 65K\). For the second interval, \(\frac{20}{t'} = 50K\). Dividing these equations yields \(t' = \frac{13}{5}t\).

Using Newton's law of cooling: \(\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_0 \right]\). For the first case, \(\frac{30}{6} = K[65 - 20] ⇒ 5 = 45K ⇒ K = \frac{1}{9}\). For the second case, \(\frac{20}{t} = \frac{1}{9}[50 - 20] = \frac{30}{9} = \frac{10}{3} ⇒ t = 6 \text{ minutes}\).

Assertion (A) correctly states Newton's law of cooling. Reason (R) is false; Newton's law of cooling is primarily derived from convection, and it is a good approximation for small temperature differences involving conduction and convection, not exclusively radiation.

Assertion (A) is the correct statement of Newton's Law of Cooling. Reason (R) is false, as the law applies to convection and can be approximated from radiation for small temperature differences, not exclusively for radiation.

Newton's law of cooling states that the rate of cooling `\(\frac{dT}{dt}\) ` is proportional to `\((T - T_s)\)`. For the first interval, average `\(T_{avg1} = 79^0 C\) ⇒ \(T_{avg1} - T_s) = 59^0 C\)`. For the second interval, average `\(T_{avg2} = 49^0 C\) ⇒ (T_{avg2} - T_s) = 29^0 C\)`. Since the temperature difference is greater in the first case, the rate of cooling is faster, meaning `\(t_1 < t_2\)`. So, Assertion (A) is false. Reason (R) states 'depends *only* on the difference', which is misleading as the rate also depends on factors like surface area and emissivity, embedded in the constant of proportionality. Thus, Reason (R) is also false under strict interpretation.