A solid body X of heat capacity C is kept in an atmosphere

whose temperature is T_A=300K. at time t=0, the temp. of X is T_O=400K. It cools according to Newton?s law of cooling. At time t₁ its temp. is found to be 350K. at this time t_1, the body X is connected to a larger body Y at atmospheric temp. T_A, through a conducting rod of length L , cross-sectional area A and thermal conductivity K. the heat capacity of Y is so large that any variation in its temp. may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. find the temperature of X at time t=3 t_1.

To solve this problem, we need to analyze the cooling of body X using Newton's law of cooling and then consider the heat transfer between body X and the larger body Y. Let's break this down step by step.

Understanding Newton's Law of Cooling

Newton's law of cooling states that the rate of heat loss of a body is proportional to the difference in temperature between the body and its surroundings. Mathematically, this can be expressed as:

Q = hA(T - T_A)

Where:

• Q = heat lost per unit time
• h = heat transfer coefficient
• A = surface area of the body
• T = temperature of the body
• T_A = ambient temperature

Cooling of Body X

Initially, body X has a temperature of T_O = 400 K. At time t = t_1, its temperature drops to T_1 = 350 K. We can use the formula for Newton's law of cooling to find the cooling constant (k):

T(t) = T_A + (T_O - T_A)e^{-kt}

Substituting the known values at time t_1:

350 = 300 + (400 - 300)e^{-kt_1}

This simplifies to:

50 = 100e^{-kt_1}

From this, we can find:

e^{-kt_1} = 0.5

Taking the natural logarithm:

-kt_1 = ln(0.5)

Thus, we have:

k = -\frac{ln(0.5)}{t_1}

Heat Transfer to Body Y

At time t_1, body X is connected to body Y through a conducting rod. Since the heat capacity of Y is very large, we can assume that its temperature remains constant at T_A = 300 K. The heat transfer between X and Y can be described by Fourier's law of heat conduction:

Q = \frac{KA(T_X - T_Y)}{L}

Where:

• K = thermal conductivity of the rod
• T_X = temperature of body X
• T_Y = temperature of body Y (constant at T_A)
• L = length of the rod

Finding Temperature at Time t = 3t_1

At time t = 3t_1, we need to consider the cooling of body X and the heat transfer to body Y. The temperature of X at this time can be expressed as:

T(3t_1) = T_A + (T_1 - T_A)e^{-k(3t_1)}

Substituting the known values:

T(3t_1) = 300 + (350 - 300)e^{-k(3t_1)}

We already know that:

e^{-kt_1} = 0.5

Thus:

e^{-k(3t_1)} = (e^{-kt_1})^3 = (0.5)^3 = 0.125

Now substituting this back into the temperature equation:

T(3t_1) = 300 + 50 \cdot 0.125

This gives:

T(3t_1) = 300 + 6.25 = 306.25 K

Final Result

Therefore, the temperature of body X at time t = 3t_1 is approximately 306.25 K.