A body cools from `61^0 C` to `59^0 C` in `T_0`. How much time it would take to cool from `51^0C` to `49^0 C`. If room temperature is `30^0 C`

Answer

To solve the problem of how long it takes for a body to cool from \(51^\circ C\) to \(49^\circ C\) given that it cools from \(61^\circ C\) to \(59^\circ C\) in time \(T_0\) with a room temperature of \(30^\circ C\), we can use Newton's Law of Cooling.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the body cools from \(61^\circ C\) to \(59^\circ C\) in time \(T_0\). We need to find the time required to cool from \(51^\circ C\) to \(49^\circ C\).

2. **Applying Newton's Law of Cooling**:
According to Newton's Law of Cooling, the rate of change of temperature of the body is proportional to the difference between its temperature and the ambient temperature (room temperature).
\[ \frac{\Delta \theta}{\Delta t} \propto (T - T_0) \]
where \(T_0\) is the ambient temperature.

3. **Calculating the Change in Temperature for the First Case**:
- Initial temperature \(T_1 = 61^\circ C\)
- Final temperature \(T_2 = 59^\circ C\)
- Change in temperature \(\Delta \theta_1 = T_1 - T_2 = 61 - 59 = 2^\circ C\)
- Average temperature \(T_{avg1} = \frac{T_1 + T_2}{2} = \frac{61 + 59}{2} = 60^\circ C\)
- The difference from room temperature \(T_{avg1} - T_0 = 60 - 30 = 30^\circ C\)

4. **Setting Up the Equation for the First Case**:
Using the proportionality:
\[ \frac{2}{T_0} = k \cdot (60 - 30) = 30k \]
This gives us:
\[ \frac{2}{T_0} = 30k \quad \text{(Equation 1)} \]

5. **Calculating the Change in Temperature for the Second Case**:
- Initial temperature \(T_3 = 51^\circ C\)
- Final temperature \(T_4 = 49^\circ C\)
- Change in temperature \(\Delta \theta_2 = T_3 - T_4 = 51 - 49 = 2^\circ C\)
- Average temperature \(T_{avg2} = \frac{T_3 + T_4}{2} = \frac{51 + 49}{2} = 50^\circ C\)
- The difference from room temperature \(T_{avg2} - T_0 = 50 - 30 = 20^\circ C\)

6. **Setting Up the Equation for the Second Case**:
Using the proportionality:
\[ \frac{2}{\Delta t} = k \cdot (50 - 30) = 20k \]
This gives us:
\[ \frac{2}{\Delta t} = 20k \quad \text{(Equation 2)} \]

7. **Relating the Two Equations**:
From Equation 1:
\[ k = \frac{2}{30T_0} \]
Substitute \(k\) into Equation 2:
\[ \frac{2}{\Delta t} = 20 \cdot \frac{2}{30T_0} \]
Simplifying gives:
\[ \frac{2}{\Delta t} = \frac{40}{30T_0} \]
Rearranging gives:
\[ \Delta t = \frac{30T_0}{20} = \frac{3}{2} T_0 \]

### Final Answer:
The time it would take to cool from \(51^\circ C\) to \(49^\circ C\) is \(\frac{3}{2} T_0\).