Question

If two body is connected by a rod of conductivity k area A length L temperature of one body is fixed and temperature of other body is not fixed mass of other body is m specific heat is s density is p the find the temperature of other body as a function of time

Answer 1

T(t) = T_{fixed} + (T_0 - T_{fixed}) e^{-\frac{kA}{mLs} t}

Answer 2

The problem involves heat transfer by conduction through a rod and the subsequent change in temperature of a body due to this heat transfer.

1. Rate of Heat Transfer by Conduction: The rate of heat flow ($dQ/dt$) through the rod connecting the two bodies is given by Fourier's Law of Conduction: $\frac{dQ}{dt} = \frac{kA}{L} (T_{fixed} - T(t))$ where:

• $k$ is the thermal conductivity of the rod.
• $A$ is the cross-sectional area of the rod.
• $L$ is the length of the rod.
• $T_{fixed}$ is the constant temperature of the first body.
• $T(t)$ is the temperature of the second body at time $t$.

2. Rate of Change of Internal Energy of the Second Body: The heat absorbed or released by the second body changes its temperature. The rate of change of heat content of the second body is given by: $\frac{dQ}{dt} = m s \frac{dT}{dt}$ where:

• $m$ is the mass of the second body.
• $s$ is the specific heat capacity of the second body.
• $dT/dt$ is the rate of change of temperature of the second body.

3. Equating the Rates and Forming a Differential Equation: By equating the two expressions for $dQ/dt$: $m s \frac{dT}{dt} = \frac{kA}{L} (T_{fixed} - T(t))$ Rearranging this equation to separate variables: $\frac{dT}{T_{fixed} - T(t)} = \frac{kA}{mLs} dt$

4. Solving the Differential Equation: Let $T_0$ be the initial temperature of the second body at $t=0$. Integrate both sides of the differential equation: $\int_{T_0}^{T(t)} \frac{dT'}{T_{fixed} - T'} = \int_0^t \frac{kA}{mLs} dt'$ To integrate the left side, let $u = T_{fixed} - T'$. Then $du = -dT'$. When $T' = T_0$, $u = T_{fixed} - T_0$. When $T' = T(t)$, $u = T_{fixed} - T(t)$. $\int_{T_{fixed} - T_0}^{T_{fixed} - T(t)} \frac{-du}{u} = \frac{kA}{mLs} t$ $-\left[\ln|u|\right]_{T_{fixed} - T_0}^{T_{fixed} - T(t)} = \frac{kA}{mLs} t$ $-\left(\ln|T_{fixed} - T(t)| - \ln|T_{fixed} - T_0|\right) = \frac{kA}{mLs} t$ $\ln\left|\frac{T_{fixed} - T_0}{T_{fixed} - T(t)}\right| = \frac{kA}{mLs} t$ Exponentiating both sides: $\frac{T_{fixed} - T_0}{T_{fixed} - T(t)} = e^{\frac{kA}{mLs} t}$ Rearranging to solve for $T(t)$: $T_{fixed} - T(t) = (T_{fixed} - T_0) e^{-\frac{kA}{mLs} t}$ $T(t) = T_{fixed} - (T_{fixed} - T_0) e^{-\frac{kA}{mLs} t}$ This can also be written as: $T(t) = T_{fixed} + (T_0 - T_{fixed}) e^{-\frac{kA}{mLs} t}$ This equation gives the temperature of the second body as a function of time. The term $\frac{mLs}{kA}$ is the thermal time constant of the system. The density $p$ given in the problem statement is not directly used, as the mass $m$ is already provided.

The final answer is $\boxed{T(t) = T_{fixed} + (T_0 - T_{fixed}) e^{-\frac{kA}{mLs} t}}$

Explanation of the solution: The heat conducted through the rod per unit time ($\frac{kA}{L} (T_{fixed} - T(t))$) is absorbed by the second body, causing its temperature to change ($m s \frac{dT}{dt}$). Equating these rates yields a first-order differential equation. Solving this separable differential equation with an initial temperature $T_0$ at $t=0$ gives the exponential decay/growth of the second body's temperature towards the fixed temperature of the first body.

Answer: The temperature of the other body as a function of time is given by: $T(t) = T_{fixed} + (T_0 - T_{fixed}) e^{-\frac{kA}{mLs} t}$ where $T_0$ is the initial temperature of the second body at $t=0$.