A solid body X of heat capacity C is kept in an atmosphere whose temperature is `T_A=300K`. At time `t=0` the temperature of X is `T_0=400K`. It cools according to Newton's law of cooling. At time `t_1`, its temperature is found to be 350K. At this time `(t_1)`, the body X is connected to a large box Y at atmospheric temperature is `T_4`, through a conducting rod of length L, cross-sectional area A and thermal conductivity K. The heat capacity Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of hte connecting rod is small compared to the surface area of X. Find the temperature of X at time `t=3t_1.`

Explanation

To solve the problem step by step, we will follow the principles of Newton's law of cooling and the conduction of heat through the connecting rod.

### Step 1: Understand the cooling process of body X
At time \( t = 0 \), the temperature of body X is \( T_0 = 400 \, K \). 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 \( T_A \).

The equation can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_A) \]
where \( k \) is a constant of proportionality.

### Step 2: Find the temperature at time \( t_1 \)
At time \( t_1 \), the temperature of body X is \( T_1 = 350 \, K \). We can use the cooling equation to relate \( T_0 \), \( T_1 \), and \( T_A \):

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

Substituting the known values:
\[ 350 = 300 + (400 - 300)e^{-kt_1} \]
\[ 350 - 300 = 100e^{-kt_1} \]
\[ 50 = 100e^{-kt_1} \]
\[ e^{-kt_1} = \frac{1}{2} \]

### Step 3: Relate the temperature drop to the time
From the above, we can express \( kt_1 \):
\[ kt_1 = \ln(2) \]

### Step 4: Connect body X to box Y
At time \( t_1 \), body X is connected to a large box Y at temperature \( T_4 \) (which is also \( T_A = 300 \, K \)). The heat loss through conduction can be expressed as:
\[ \frac{dQ}{dt} = \frac{KA(T - T_4)}{L} \]
where \( K \) is the thermal conductivity, \( A \) is the cross-sectional area, and \( L \) is the length of the rod.

### Step 5: Set up the differential equation for the new cooling process
Now, we can write the new rate of temperature change for body X:
\[ C\frac{dT}{dt} = -\frac{KA(T - T_A)}{L} \]
This leads to:
\[ \frac{dT}{dt} = -\frac{KA}{CL}(T - T_A) \]

### Step 6: Solve the new differential equation
Integrating this equation gives:
\[ T(t) = T_A + (T_1 - T_A)e^{-\frac{KA}{CL}(t - t_1)} \]
Substituting \( T_A = 300 \, K \) and \( T_1 = 350 \, K \):
\[ T(t) = 300 + (350 - 300)e^{-\frac{KA}{CL}(t - t_1)} \]
\[ T(t) = 300 + 50e^{-\frac{KA}{CL}(t - t_1)} \]

### Step 7: Find the temperature at \( t = 3t_1 \)
Now substituting \( t = 3t_1 \):
\[ T(3t_1) = 300 + 50e^{-\frac{KA}{CL}(3t_1 - t_1)} \]
\[ T(3t_1) = 300 + 50e^{-\frac{KA}{CL}(2t_1)} \]

### Step 8: Substitute \( kt_1 \) into the equation
Since we found that \( kt_1 = \ln(2) \):
\[ T(3t_1) = 300 + 50e^{-2\frac{KA}{CL}\frac{\ln(2)}{k}} \]
This can be simplified further based on the specific values of \( K, A, C, L \).

### Final Result
Thus, the temperature of body X at time \( t = 3t_1 \) can be expressed as:
\[ T(3t_1) = 300 + 50e^{-2\frac{KA}{CL}\frac{\ln(2)}{k}} \]