Let's assume that the temperature at a point x cm from the high temperature end is T(x). We know that the temperature gradient in the rod is proportional to the rate of heat flow through the rod, which is given by Fourier's law of heat conduction:
q = -kA(dT/dx)
where q is the rate of heat flow, k is the thermal conductivity of copper, A is the cross-sectional area of the rod, and (dT/dx) is the temperature gradient.
Since the rod is in steady state, the rate of heat flow is constant throughout the rod, so we can write:
q = kA(dT/dx) = constant
Integrating both sides with respect to x, we get:
∫(dT/dx)dx = (1/kA)∫qdx
Integrating from 0 to L (where L is the length of the rod), we get:
T(L) - T(0) = (1/kA)qL
Since T(0) = 0°C and T(L) = 20°C, we can write:
20 - 0 = (1/kA)q(60)
Simplifying, we get:
q = 1/3 kA
Now, let's consider a point 20 cm from the high temperature end. We know that the rate of heat flow through the rod is constant, so we can write:
q = -kA(dT/dx)
Substituting the given values, we get:
1/3 kA = -kA (dT/dx) at x = 20
Simplifying, we get:
(dT/dx) = -1/3
Integrating both sides with respect to x, we get:
T(x) = (-1/3)x + C
where C is the constant of integration.
Using the boundary condition that T(0) = 0°C, we get:
C = 0
Therefore, the temperature at a point 20 cm from the high temperature end is:
T(20) = (-1/3)(20) + 20 = 13.3°C
So, the temperature at a point 20cm from the high temperature end is 13.3°C.