A long iron rod initially at a temperature of 20°C has one end dipped in boiling water (100°C) at time, t = 0. The curved surface of the rod is insulated so that heat conduction is one dimensional in the axial direction. The temperature at a distance 100 mm from the dipped end becomes 40°C at time, t = 200 s. The same temperature is achieved at a distance of 200 mm from the dipped end at time

ANS:D - t = 800 s

To solve this problem, we can use the one-dimensional heat conduction equation in cylindrical coordinates: ∂x2∂2T​=α1​∂t∂T​ where T is the temperature, x is the distance along the rod, and α is the thermal diffusivity of the material. Given that the rod is initially at 20°C and one end is dipped in boiling water at 100°C, we can define the boundary condition at =0x=0 as (0,)=100T(0,t)=100 and at =x=L (the length of the rod) as (,)=20T(L,t)=20. We can solve this partial differential equation subject to these boundary conditions to find the temperature distribution along the rod as a function of time. Once we have the temperature distribution, we can determine the time at which the temperature at a distance of 200 mm from the dipped end reaches 40°C. Given that the temperature at a distance of 100 mm from the dipped end becomes 40°C at =200t=200 s, we can use this information to estimate the thermal diffusivity (α) using the formula: =24⋅α=4⋅td2​ where d is the diameter of the rod (assuming the rod is cylindrical). Let's calculate α: =(0.1)24⋅200=1.25×10−4 m2/sα=4⋅200(0.1)2​=1.25×10−4m2/s Now, using this value of α, we can solve the heat conduction equation to find the temperature distribution along the rod and determine the time at which the temperature at a distance of 200 mm from the dipped end reaches 40°C. Let's solve the equation to find the time. To solve the problem, we first need to find the thermal diffusivity (α) of iron using the provided information: =24⋅α=4⋅td2​ Given that the distance from the dipped end to where the temperature is 40°C is 100 mm (0.1 m), and the time taken for this temperature to be reached is 200 s, we can calculate: =(0.1)24⋅200=1.25×10−4 m2/sα=4⋅200(0.1)2​=1.25×10−4m2/s Now, using the one-dimensional heat conduction equation: ∂2∂2=1∂∂∂x2∂2T​=α1​∂t∂T​ We can solve this equation subject to the boundary conditions: At =0x=0: (0)=100T(0,t)=100 At =x=L: (,)=20T(L,t)=20 We'll use the method of separation of variables and apply appropriate boundary conditions to find the temperature distribution (,)T(x,t) along the rod. After obtaining the temperature distribution, we'll find the time t at which the temperature at a distance of 200 mm from the dipped end reaches 40°C. Let's solve the heat conduction equation to find the time. Given the thermal diffusivity =1.25×10−4 m2/sα=1.25×10−4m2/s, and the boundary conditions, we'll solve the one-dimensional heat conduction equation using separation of variables. Let's denote (,)=⋅()T(x,t)=X(x)⋅T(t). Then the heat conduction equation becomes: 1​dx2d2X​=α1​T1​dtdT​ Since the left side of the equation depends only on x and the right side depends only on t, they must be equal to a constant, denoted by −2−λ2, to satisfy the equality for all x and t. This leads to two ordinary differential equations:

1. For ()X(x):

2+2=0dx2d2X​+λ2X=0

2. For ()T(t):

=0dtdT​+αλ2T=0 The general solutions to these differential equations are:

1. For ()X(x):

X(x)=Acos(λx)+Bsin(λx)

2. For T(t):

T(t)=Ce−αλ2t We apply the boundary conditions: At =0x=0: (0,)=100T(0,t)=100 At =x=L:)=20T(L,t)=20 Applying the boundary condition at =0x=0, we have: =100T(0,t)=X(0)⋅T(t)=(A⋅1)⋅T(t)=A⋅T(t)=100 Similarly, applying the boundary condition at x=L, we have: 20T(L,t)=X(L)⋅T(t)=(Acos(λL)+Bsin(λL))⋅T(t)=20 To satisfy this boundary condition, =100A=100 and cos⁡()=0cos(λL)=0, which implies λL=2π​+nπ where n is an integer. So, =(2+1)2λ=2L(2n+1)π​. Now, we apply the initial condition: ,0)=20T(x,0)=20. This implies X(x) is constant, so we take=0B=0 (no sin term). Finally, we substitute λ into the equation for T(t), and )=20T(t)=20 when =0x=0 and =200t=200. Substituting the values, we can solve forn, and then find the time t at which =40T(x,t)=40 when =200x=200. Let's solve for t. Given =100cos⁡X(x)=100cos(λx), the temperature distribution along the rod is: T(x,t)=100cos(2L(2n+1)πx​)e−α(2L(2n+1)π​)2t At =0t=0, we have ,0)=20T(x,0)=20. Substituting into the equation, we find: =20T(x,0)=100cos(2L(2n+1)πx​)=20 This implies =0n=0 (to satisfy the cosine term). Soλ=2Lπ​, and the temperature distribution along the rod is: T(x,t)=100cos(2Lπx​)e−α(2Lπ​)2t We need to find the time t at which (0.2=40T(0.2,t)=40. Substituting =0.2x=0.2 and =40T=40 into the equation, we get: 40=100cos⁡40=100cos(2Lπ⋅0.2​)e−α(2Lπ​)2t Solving this equation for t, we can find the time at which =40T=40 when =0.2x=0.2. Let's solve for t. To find the time =40T(0.2,t)=40, we can rearrange the equation: 40=100cos⁡40=100cos(2Lπ⋅0.2​)e−α(2Lπ​)2t cos⁡=40100cos(2Lπ⋅0.2​)=10040​ cos⁡=0.4cos(2Lπ​)=0.4 To find the time t, we take the natural logarithm of both sides: ln⁡(cos⁡ln(cos(2Lπ​))=ln(0.4)−α(2Lπ​)2t Given that =1.25×10−4 m2/sα=1.25×10−4m2/s and =0.2L=0.2, we can substitute these values into the equation to find t. ln⁡(cos⁡ln(cos(0.4π​))=ln(0.4)−1.25×10−4(0.4π​)2t ln⁡(cos⁡(ln(cos(45π​))=ln(0.4)−1.25×10−4(0.4π​)2t ln⁡(cos⁡(4))=ln⁡(0.4)−1.25×10−4(0.4)2ln(cos(4π​))=ln(0.4)−1.25×10−4(0.4π​)2t ln⁡(cos⁡(4))=ln⁡(0.4)−1.25×10−420.42ln(cos(4π​))=ln(0.4)−0.421.25×10−4π2​t