Certain quantity of water cools from `70^(@)C` to `60^(@)C` in the first `5` minutes and to `54^(@)C` in the next `5` minutes. The temperature of the surrounding is

Options

A. `45^(@)C`
B. `20^(@)C`
C. `42^(@)C`
D. `10^(@)C`

Explanation

To solve the problem of determining the surrounding temperature using Newton's Law of Cooling, we can follow these steps:

### Step 1: Understand the cooling process
We know that the water cools from 70°C to 60°C in the first 5 minutes, and then from 60°C to 54°C in the next 5 minutes. We need to find the surrounding temperature (θ₀).

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings. This can be expressed mathematically as:

\[ \frac{\Delta T}{\Delta t} = k (T - T_0) \]

Where:
- \( \Delta T \) is the change in temperature,
- \( \Delta t \) is the time interval,
- \( k \) is a constant,
- \( T \) is the temperature of the object,
- \( T_0 \) is the surrounding temperature.

### Step 3: Set up the equations for the two intervals
1. For the first interval (from 70°C to 60°C in 5 minutes):
- Initial temperature (T₁) = 70°C
- Final temperature (T₂) = 60°C
- Average temperature (T_avg1) = (70 + 60) / 2 = 65°C

The equation becomes:
\[ \frac{70 - 60}{5} = k (65 - T_0) \]
Simplifying gives:
\[ 2 = k (65 - T_0) \quad \text{(Equation 1)} \]

2. For the second interval (from 60°C to 54°C in the next 5 minutes):
- Initial temperature (T₁) = 60°C
- Final temperature (T₂) = 54°C
- Average temperature (T_avg2) = (60 + 54) / 2 = 57°C

The equation becomes:
\[ \frac{60 - 54}{5} = k (57 - T_0) \]
Simplifying gives:
\[ \frac{6}{5} = k (57 - T_0) \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( 2 = k (65 - T_0) \)
2. \( \frac{6}{5} = k (57 - T_0) \)

We can express \( k \) from both equations:
From Equation 1:
\[ k = \frac{2}{65 - T_0} \]
From Equation 2:
\[ k = \frac{6/5}{57 - T_0} \]

Setting these equal to each other:
\[ \frac{2}{65 - T_0} = \frac{6/5}{57 - T_0} \]

### Step 5: Cross-multiply and solve for T₀
Cross-multiplying gives:
\[ 2(57 - T_0) = \frac{6}{5}(65 - T_0) \]

Expanding both sides:
\[ 114 - 2T_0 = \frac{390 - 6T_0}{5} \]

Multiplying through by 5 to eliminate the fraction:
\[ 570 - 10T_0 = 390 - 6T_0 \]

Rearranging gives:
\[ 570 - 390 = 10T_0 - 6T_0 \]
\[ 180 = 4T_0 \]

Dividing by 4:
\[ T_0 = 45°C \]

### Final Answer
The temperature of the surrounding is **45°C**.

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