A body cools down from `50^(@)C` to `45^(@)C` in 5 minutes and to `40^(@)C` in another 8 minutes. Find the temperature of the surrounding.

Explanation

To find the temperature of the surrounding (T₀) using the information provided, we can follow these steps:

### Step 1: Setup the Problem
We know that a body cools from 50°C to 45°C in 5 minutes and then from 45°C to 40°C in another 8 minutes. We need to find the temperature of the surrounding (T₀).

### Step 2: Calculate Average Temperatures
1. For the first cooling interval (from 50°C to 45°C):
- Average temperature (T_avg1) = (50 + 45) / 2 = 47.5°C
2. For the second cooling interval (from 45°C to 40°C):
- Average temperature (T_avg2) = (45 + 40) / 2 = 42.5°C

### Step 3: Calculate the Rate of Cooling
1. For the first interval:
- Change in temperature (ΔT1) = 50°C - 45°C = 5°C
- Time (t1) = 5 minutes
- Rate of cooling (R1) = ΔT1 / t1 = 5°C / 5 min = 1°C/min
2. For the second interval:
- Change in temperature (ΔT2) = 45°C - 40°C = 5°C
- Time (t2) = 8 minutes
- Rate of cooling (R2) = ΔT2 / t2 = 5°C / 8 min = 0.625°C/min

### Step 4: Apply Newton's Law of Cooling
According to Newton's Law of Cooling:
- Rate of change of temperature (dT/dt) = -k(T - T₀)

For the first interval:
- R1 = -k(T_avg1 - T₀)
- 1 = -k(47.5 - T₀) (Equation 1)

For the second interval:
- R2 = -k(T_avg2 - T₀)
- 0.625 = -k(42.5 - T₀) (Equation 2)

### Step 5: Divide the Equations
We can divide Equation 1 by Equation 2 to eliminate k:
\[ \frac{1}{0.625} = \frac{-(47.5 - T₀)}{-(42.5 - T₀)} \]

### Step 6: Simplify the Equation
1. Calculate the left side:
- \( \frac{1}{0.625} = 1.6 \)
2. Set up the equation:
- \( 1.6 = \frac{47.5 - T₀}{42.5 - T₀} \)

### Step 7: Cross-Multiply and Solve for T₀
Cross-multiplying gives:
\[ 1.6(42.5 - T₀) = 47.5 - T₀ \]
Expanding this:
\[ 68 - 1.6T₀ = 47.5 - T₀ \]
Rearranging gives:
\[ 68 - 47.5 = 1.6T₀ - T₀ \]
\[ 20.5 = 0.6T₀ \]
\[ T₀ = \frac{20.5}{0.6} \]
\[ T₀ = 34.17°C \]

### Conclusion
The temperature of the surrounding (T₀) is approximately **34°C**.