A hot body placed in air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being `k` times the temperature difference from the surrounding. Starting from `t=0` , find the time in which the body will lose half the maximum heat it can lose.

Explanation

To solve the problem of finding the time in which a hot body loses half the maximum heat it can lose according to Newton's law of cooling, we can follow these steps:

### Step 1: Understand Newton's Law of Cooling
According to Newton's law of cooling, the rate of change of temperature of a body is proportional to the difference between its temperature and the ambient temperature. Mathematically, this can be expressed as:

\[ \frac{dT}{dt} = -k(T - T_0) \]

where:
- \( T \) is the temperature of the body at time \( t \),
- \( T_0 \) is the ambient temperature,
- \( k \) is a positive constant.

### Step 2: Set Up the Initial Conditions
Let \( T_1 \) be the initial temperature of the body at \( t = 0 \). Therefore, we have:
- At \( t = 0 \), \( T = T_1 \).

### Step 3: Integrate the Equation
We can separate variables and integrate:

\[ \int_{T_1}^{T} \frac{dT}{T - T_0} = -k \int_{0}^{t} dt \]

This gives us:

\[ \ln |T - T_0| \bigg|_{T_1}^{T} = -kt \]

### Step 4: Solve the Integration
After integrating, we can express this as:

\[ \ln |T - T_0| - \ln |T_1 - T_0| = -kt \]

This simplifies to:

\[ \ln \left(\frac{T - T_0}{T_1 - T_0}\right) = -kt \]

### Step 5: Exponentiate to Remove the Logarithm
Exponentiating both sides gives:

\[ \frac{T - T_0}{T_1 - T_0} = e^{-kt} \]

Rearranging this, we find:

\[ T = T_0 + (T_1 - T_0)e^{-kt} \]

### Step 6: Determine the Maximum Heat Loss
The maximum heat loss \( Q \) that the body can lose is given by:

\[ Q = mc(T_1 - T_0) \]

To find the time when the body loses half of this maximum heat, we need to find when the temperature decreases by half of the maximum temperature difference:

\[ \Delta T = \frac{Q}{2} = \frac{mc(T_1 - T_0)}{2} \]

### Step 7: Set Up the Equation for Half Heat Loss
The temperature at which half the maximum heat is lost can be expressed as:

\[ T = T_0 + \frac{(T_1 - T_0)}{2} \]

### Step 8: Substitute and Solve for Time
Substituting this value of \( T \) into our equation from Step 5:

\[ \frac{T_0 + \frac{(T_1 - T_0)}{2} - T_0}{T_1 - T_0} = e^{-kt} \]

This simplifies to:

\[ \frac{(T_1 - T_0)/2}{T_1 - T_0} = e^{-kt} \]

Thus:

\[ \frac{1}{2} = e^{-kt} \]

Taking the natural logarithm of both sides gives:

\[ \ln\left(\frac{1}{2}\right) = -kt \]

### Step 9: Solve for Time \( t \)
Rearranging this gives:

\[ t = -\frac{\ln(1/2)}{k} = \frac{\ln(2)}{k} \]

### Final Answer
The time in which the body will lose half the maximum heat it can lose is:

\[ t = \frac{\ln(2)}{k} \]