If a body cools down from `80^(@) C`to `60^(@) C` in 10 min when the temperature of the surrounding of the is `30^(@) C` . Then, the temperature of the body after next 10 min will be

Options

A. `50^(@) C`
B. `48^(@) C`
C. `30^(@) C`
D. None of the above

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of a body is proportional to the difference between its own temperature and the surrounding temperature.

### Step 1: Set up the initial conditions
We know that the initial temperature of the body (T1) is 80°C, and it cools down to a final temperature (T2) of 60°C in 10 minutes. The surrounding temperature (T0) is 30°C.

### Step 2: Apply Newton's Law of Cooling for the first interval
According to Newton's Law of Cooling:
\[ \frac{T1 - T2}{t} = K \left( \frac{T1 + T2}{2} - T0 \right) \]
Substituting the known values:
\[ \frac{80 - 60}{10} = K \left( \frac{80 + 60}{2} - 30 \right) \]
This simplifies to:
\[ \frac{20}{10} = K \left( 70 - 30 \right) \]
\[ 2 = K \cdot 40 \]
Thus, we find:
\[ K = \frac{2}{40} = \frac{1}{20} \]

### Step 3: Set up the equation for the next interval
Now, we need to find the temperature of the body after the next 10 minutes. Let the temperature after the next 10 minutes be T (unknown). We can apply Newton's Law of Cooling again:
\[ \frac{T2 - T}{10} = K \left( \frac{T2 + T}{2} - T0 \right) \]
Substituting T2 = 60°C and K = \frac{1}{20}:
\[ \frac{60 - T}{10} = \frac{1}{20} \left( \frac{60 + T}{2} - 30 \right) \]
This simplifies to:
\[ \frac{60 - T}{10} = \frac{1}{20} \left( \frac{60 + T - 60}{2} \right) \]
\[ \frac{60 - T}{10} = \frac{1}{20} \left( \frac{T}{2} \right) \]
\[ \frac{60 - T}{10} = \frac{T}{40} \]

### Step 4: Cross-multiply to solve for T
Cross-multiplying gives:
\[ 40(60 - T) = 10T \]
Expanding this:
\[ 2400 - 40T = 10T \]
Combining like terms:
\[ 2400 = 50T \]
Thus:
\[ T = \frac{2400}{50} = 48°C \]

### Conclusion
The temperature of the body after the next 10 minutes will be **48°C**.

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