A body cools down from `45^(@)C` to `40^(@)C` in 5 minutes and to `35^(@)` in another 8 minutes. Find the temperature of the surrounding.

Answer

To find the temperature of the surrounding using the information given in the problem, we can apply Newton's Law of Cooling. Here’s a step-by-step solution:

### Step 1: Define the Variables
Let \( T \) be the temperature of the surroundings. We know:
- Initial temperature \( T_1 = 45^\circ C \)
- Final temperature after 5 minutes \( T_2 = 40^\circ C \)
- Final temperature after another 8 minutes \( T_3 = 35^\circ C \)

### Step 2: Calculate the Average Temperature for Case 1
For the first cooling period (from \( 45^\circ C \) to \( 40^\circ C \)):
- Average temperature \( T_{avg1} = \frac{T_1 + T_2}{2} = \frac{45 + 40}{2} = 42.5^\circ C \)

### Step 3: Calculate the Rate of Cooling for Case 1
The rate of cooling can be calculated as:
\[ \text{Rate of cooling} = \frac{T_1 - T_2}{\text{Time}} = \frac{45 - 40}{5} = 1^\circ C/\text{min} \]

According to Newton's Law of Cooling:
\[ \frac{dT}{dt} = -k (T_{avg1} - T) \]
Where \( k \) is a constant. Thus, we can express \( k \) as:
\[ k = \frac{1}{42.5 - T} \]

### Step 4: Calculate the Average Temperature for Case 2
For the second cooling period (from \( 40^\circ C \) to \( 35^\circ C \)):
- Average temperature \( T_{avg2} = \frac{T_2 + T_3}{2} = \frac{40 + 35}{2} = 37.5^\circ C \)

### Step 5: Calculate the Rate of Cooling for Case 2
The rate of cooling for the second period is:
\[ \text{Rate of cooling} = \frac{T_2 - T_3}{\text{Time}} = \frac{40 - 35}{8} = \frac{5}{8}^\circ C/\text{min} \]

Using Newton's Law of Cooling again:
\[ \frac{dT}{dt} = -k (T_{avg2} - T) \]
Thus, we can express \( k \) for the second case as:
\[ k = \frac{5/8}{37.5 - T} \]

### Step 6: Set the Two Expressions for \( k \) Equal
Since \( k \) is the same in both cases, we can set the two equations equal to each other:
\[ \frac{1}{42.5 - T} = \frac{5/8}{37.5 - T} \]

### Step 7: Cross Multiply and Solve for \( T \)
Cross multiplying gives:
\[ 8 = 5 \cdot \frac{(42.5 - T)}{(37.5 - T)} \]
This simplifies to:
\[ 8(37.5 - T) = 5(42.5 - T) \]
Expanding both sides:
\[ 300 - 8T = 212.5 - 5T \]
Rearranging gives:
\[ 300 - 212.5 = 8T - 5T \]
\[ 87.5 = 3T \]
Thus:
\[ T = \frac{87.5}{3} \approx 29.17^\circ C \]

### Final Answer
The temperature of the surrounding is approximately \( 29.17^\circ C \).

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