Hot water cools from `60^@C` to `50^@C` in the first 10 min and to `42^@C` in the next 10 min. The temperature of the surrounding is

Options

A. `5^@C`
B. `10^@C`
C. `15^@C`
D. `20^@C`

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature (surrounding temperature).

Let's denote:
- \( T_1 = 60^\circ C \) (initial temperature of the water),
- \( T_2 = 50^\circ C \) (temperature after 10 minutes),
- \( T_3 = 42^\circ C \) (temperature after another 10 minutes),
- \( T_0 \) = surrounding temperature (which we need to find).

### Step 1: Apply Newton's Law of Cooling for the first interval (from 60°C to 50°C)

Using the formula:
\[ \frac{-dT}{dt} = k(T - T_0) \]

For the first interval:
- Change in temperature, \( dT = T_2 - T_1 = 50 - 60 = -10 \)
- Time, \( dt = 10 \) minutes.

The average temperature during this interval is:
\[ T_{avg1} = \frac{T_1 + T_2}{2} = \frac{60 + 50}{2} = 55^\circ C \]

Substituting into Newton's Law:
\[ \frac{-(-10)}{10} = k(55 - T_0) \]
\[ 1 = k(55 - T_0) \quad \text{(Equation 1)} \]

### Step 2: Apply Newton's Law of Cooling for the second interval (from 50°C to 42°C)

For the second interval:
- Change in temperature, \( dT = T_3 - T_2 = 42 - 50 = -8 \)
- Time, \( dt = 10 \) minutes.

The average temperature during this interval is:
\[ T_{avg2} = \frac{T_2 + T_3}{2} = \frac{50 + 42}{2} = 46^\circ C \]

Substituting into Newton's Law:
\[ \frac{-(-8)}{10} = k(46 - T_0) \]
\[ 0.8 = k(46 - T_0) \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

We have two equations:
1. \( 1 = k(55 - T_0) \)
2. \( 0.8 = k(46 - T_0) \)

From Equation 1, we can express \( k \):
\[ k = \frac{1}{55 - T_0} \]

Substituting \( k \) into Equation 2:
\[ 0.8 = \frac{1}{55 - T_0}(46 - T_0) \]

Cross-multiplying gives:
\[ 0.8(55 - T_0) = 46 - T_0 \]
\[ 44 - 0.8T_0 = 46 - T_0 \]

Rearranging terms:
\[ T_0 - 0.8T_0 = 46 - 44 \]
\[ 0.2T_0 = 2 \]
\[ T_0 = \frac{2}{0.2} = 10^\circ C \]

### Final Answer
The temperature of the surrounding is \( T_0 = 10^\circ C \).

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