Solution:

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.

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Solution:

Let the temperature of the surroundings be \( T_s \).

Given:
- First cooling: from \( 62^\circ \text{C} \) to \( 50^\circ \text{C} \) in 10 minutes.
- Second cooling: from \( 50^\circ \text{C} \) to \( 42^\circ \text{C} \) in the next 10 minutes.

Step 1: Use Newton's Law of Cooling
According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Step 2: Substitute Values
Let \( T_1 = 62^\circ \text{C} \), \( T_2 = 50^\circ \text{C} \), and \( T_3 = 42^\circ \text{C} \).

Then:
\[
\frac{62 - 50}{62 - T_s} = \frac{50 - 42}{50 - T_s}
\]

Step 3: Solve for \( T_s \)
\[
\frac{12}{62 - T_s} = \frac{8}{50 - T_s}
\]

Cross-multiplying:
\[
12(50 - T_s) = 8(62 - T_s)
\]

Expanding and simplifying:
\[
600 - 12T_s = 496 - 8T_s
\]

\[
4T_s = 104 \Rightarrow T_s = 26^\circ \text{C}
\]

So, the temperature of the surroundings is \( 26^\circ \text{C} \).

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Solution:

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.

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Solution:

Using Newton's law of cooling:

Given:
- Initial cooling: from \( T_1 = 61^\circ \text{C} \) to \( T_2 = 59^\circ \text{C} \) in \( \Delta t_1 = 10 \) minutes, with room temperature \( T_s = 30^\circ \text{C} \).
- New cooling required: from \( T_3 = 51^\circ \text{C} \) to \( T_4 = 49^\circ \text{C} \).

Using the proportionality from Newton's law:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{\text{Average Temp Difference in Initial Cooling}}{\text{Average Temp Difference in New Cooling}}
\]

Calculate average temperature differences:
1. For initial cooling: \( \frac{61 + 59}{2} - 30 = 60 - 30 = 30 \)
2. For new cooling: \( \frac{51 + 49}{2} - 30 = 50 - 30 = 20 \)

Then:
\[
\frac{10}{\Delta t_2} = \frac{30}{20}
\]

Solving for \( \Delta t_2 \):
\[
\Delta t_2 = 10 \times \frac{20}{30} = 15 \text{ minutes}
\]

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Solution:

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

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Solution:

Using Newton's law of cooling:

Given:
- Initial cooling: \( T_1 = 80^\circ \text{C} \) to \( T_2 = 79.9^\circ \text{C} \) in 5 min, surroundings \( T_s = 20^\circ \text{C} \).
- New cooling required: from \( 70^\circ \text{C} \) to \( 69.9^\circ \text{C} \).

For both cases, Newton's law states:
\[
\frac{\Delta T}{\Delta t} \propto (T - T_s)
\]

Since the temperature difference is small in both cases, we assume:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{T_1 - T_s}{T_3 - T_s}
\]

So:
\[
\frac{5}{\Delta t_2} = \frac{80 - 20}{70 - 20} \Rightarrow \Delta t_2 = 5 \times \frac{70 - 20}{80 - 20} = 5 \times \frac{50}{60} = 6 \text{ min}
\]

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Solution:

Using Newton's law of cooling:

Let \( T_s \) be the temperature of the surroundings, and \( T_1 = 70^\circ \text{C} \), \( T_2 = 60^\circ \text{C} \), \( T_3 = 54^\circ \text{C} \).

The average temperatures in each time interval are:
1. For first 5 minutes: \( \frac{70 + 60}{2} = 65^\circ \text{C} \)
2. For next 5 minutes: \( \frac{60 + 54}{2} = 57^\circ \text{C} \)

According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Substitute values:
\[
\frac{70 - 60}{65 - T_s} = \frac{60 - 54}{57 - T_s}
\]

Simplifying:
\[
\frac{10}{65 - T_s} = \frac{6}{57 - T_s}
\]

Cross-multiplying and solving for \( T_s \) gives \( T_s = 45^\circ \text{C} \).

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