Hot water cools from `60^(@)C` to `50^(@)C` in the first 10 minutes and to `45^(@)C` in the next 10 minutes. The temperature of the surroundings is:

Options

A. `25^(@)C`
B. `10^(@)C`
C. `15^(@)C`
D. `20^(@)C`

Answer

To solve the problem of determining 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).

### Step 2: Set Up the Equations
Let:
- \( T_1 = 60^\circ C \) (initial temperature of the water)
- \( T_2 = 50^\circ C \) (temperature after 10 minutes)
- \( T_3 = 45^\circ C \) (temperature after another 10 minutes)
- \( T_s \) = temperature of the surroundings (unknown)

According to Newton's law of cooling, we can write two equations based on the cooling process:

1. For the first 10 minutes (from 60°C to 50°C):
\[ \frac{dT}{dt} \propto (T - T_s) \]
This gives us:
\[ \frac{60 - T_s}{50 - T_s} = k \cdot 10 \quad (1) \]

2. For the next 10 minutes (from 50°C to 45°C):
\[ \frac{dT}{dt} \propto (T - T_s) \]
This gives us:
\[ \frac{50 - T_s}{45 - T_s} = k \cdot 10 \quad (2) \]

### Step 3: Solve the Equations
From equation (1):
\[ \frac{60 - T_s}{50 - T_s} = k \cdot 10 \]

From equation (2):
\[ \frac{50 - T_s}{45 - T_s} = k \cdot 10 \]

Now, we can set the two equations equal to each other since both are equal to \( k \cdot 10 \):
\[ \frac{60 - T_s}{50 - T_s} = \frac{50 - T_s}{45 - T_s} \]

### Step 4: Cross-Multiply and Simplify
Cross-multiplying gives us:
\[ (60 - T_s)(45 - T_s) = (50 - T_s)(50 - T_s) \]

Expanding both sides:
\[ 2700 - 105T_s + T_s^2 = 2500 - 100T_s + T_s^2 \]

### Step 5: Cancel Out and Rearrange
Cancelling \( T_s^2 \) from both sides:
\[ 2700 - 105T_s = 2500 - 100T_s \]

Rearranging gives:
\[ 2700 - 2500 = 105T_s - 100T_s \]
\[ 200 = 5T_s \]

### Step 6: Solve for \( T_s \)
Dividing both sides by 5:
\[ T_s = 40^\circ C \]

### Final Answer
The temperature of the surroundings is \( 40^\circ C \).

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