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`

Answer

To solve the problem of finding the temperature of the surroundings using Newton's law of cooling, we can follow these steps:

### Step 1: Understand Newton's Law of Cooling
Newton's law of cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature (temperature of the surroundings). The formula can be expressed as:

\[ \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right) \]

Where:
- \(T_1\) = initial temperature of the object
- \(T_2\) = final temperature of the object
- \(t\) = time interval
- \(k\) = cooling constant
- \(T_s\) = temperature of the surroundings

### Step 2: Set Up the First Equation
From the problem, the water cools from \(70^\circ C\) to \(60^\circ C\) in the first 5 minutes. We can substitute the values into the formula:

- \(T_1 = 70^\circ C\)
- \(T_2 = 60^\circ C\)
- \(t = 5\) minutes

Substituting these values into the equation gives:

\[ \frac{70 - 60}{5} = k \left( \frac{70 + 60}{2} - T_s \right) \]

This simplifies to:

\[ 2 = k \left( 65 - T_s \right) \quad \text{(Equation 1)} \]

### Step 3: Set Up the Second Equation
Next, we consider the cooling from \(60^\circ C\) to \(54^\circ C\) in the next 5 minutes. Again, we substitute the values:

- \(T_1 = 60^\circ C\)
- \(T_2 = 54^\circ C\)
- \(t = 5\) minutes

Substituting these values into the equation gives:

\[ \frac{60 - 54}{5} = k \left( \frac{60 + 54}{2} - T_s \right) \]

This simplifies to:

\[ 1.2 = k \left( 57 - T_s \right) \quad \text{(Equation 2)} \]

### Step 4: Solve the Two Equations
Now we have two equations:

1. \(2 = k(65 - T_s)\)
2. \(1.2 = k(57 - T_s)\)

We can solve these equations by dividing Equation 1 by Equation 2 to eliminate \(k\):

\[ \frac{2}{1.2} = \frac{65 - T_s}{57 - T_s} \]

This simplifies to:

\[ \frac{5}{3} = \frac{65 - T_s}{57 - T_s} \]

### Step 5: Cross Multiply and Solve for \(T_s\)
Cross-multiplying gives:

\[ 5(57 - T_s) = 3(65 - T_s) \]

Expanding both sides:

\[ 285 - 5T_s = 195 - 3T_s \]

Rearranging terms gives:

\[ 285 - 195 = 5T_s - 3T_s \]

This simplifies to:

\[ 90 = 2T_s \]

### Step 6: Calculate \(T_s\)
Dividing both sides by 2 gives:

\[ T_s = 45^\circ C \]

### Conclusion
The temperature of the surroundings is \(T_s = 45^\circ C\).

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