one end of a metal rod of length 0.3 m area 60 cm2 is in boiling water and the other end is in melting ice. if 0.12 kg of ice melts per minute, then coefficient of thermal conductivity of the material of the rod is

To find the coefficient of thermal conductivity of the metal rod, we can use the principle of heat transfer through conduction. The heat lost by the boiling water at one end of the rod is equal to the heat gained by the ice at the other end. This relationship can be expressed using the formula for heat transfer through conduction, which is given by Fourier's law:

Fourier's Law of Heat Conduction

The formula is:

Q = k * A * (T1 - T2) * t / L

• Q = heat transferred (in joules)
• k = coefficient of thermal conductivity (in W/m·K)
• A = cross-sectional area (in m²)
• T1 - T2 = temperature difference (in K)
• t = time (in seconds)
• L = length of the rod (in meters)

Given Data

From the problem, we have:

• Length of the rod, L = 0.3 m
• Area, A = 60 cm² = 60 × 10^-4 m² = 0.006 m²
• Mass of ice melted per minute, m = 0.12 kg/min
• Heat required to melt ice, Q = m * Lf, where Lf is the latent heat of fusion for ice, approximately 334,000 J/kg

Calculating Heat Transfer

First, we need to convert the mass of ice melted per minute into seconds:

m = 0.12 kg/min = 0.12 kg/60 s = 0.002 kg/s

Now, we can calculate the heat absorbed by the ice:

Q = m * Lf = 0.002 kg/s * 334,000 J/kg = 668 J/s

Temperature Difference

Next, we need to determine the temperature difference between the boiling water and melting ice. The boiling point of water is 100°C and the melting point of ice is 0°C, so:

T1 - T2 = 100°C - 0°C = 100 K

Substituting Values into Fourier's Law

Now we can rearrange Fourier's law to solve for the coefficient of thermal conductivity, k:

k = Q * L / (A * (T1 - T2) * t)

Substituting the known values:

k = 668 J/s * 0.3 m / (0.006 m² * 100 K)

k = 668 * 0.3 / (0.006 * 100)

k = 200.4 / 0.6 = 334 W/m·K

Final Result

The coefficient of thermal conductivity of the material of the rod is approximately 334 W/m·K. This value indicates how effectively the material conducts heat, which is crucial in applications involving thermal management.