Question

A slab of stone of area 0.36 m square and thickness 0.1 m is exposed on the lower surface to steam at 100 degree c.a block of ice at 0 degree c rests on the upper surface of he slab .in one hour 4.8 kg of ice is melted .the thermal conductivity of slab is latent heat of fusion of ice = 3.36 into 10 raise to power 5 ???????????????

Answer 1

Answer on Question # 37026

Physics - Thermodynamics

Question:

A slab of stone of area $0.36\,\mathrm{m}$ square and thickness $0.1\,\mathrm{m}$ is exposed on the lower surface to steam at 100 degree c.a block of ice at 0 degree c rests on the upper surface of the slab. In one hour $4.8\,\mathrm{kg}$ of ice is melted. The thermal conductivity of slab is latent heat of fusion of ice = 3.36 into 10 raise to power $5\,\mathrm{J/kg}$.

Solution:

$$S = 0.36\,m^2$$ $$d = 0.1\,\mathrm{m}$$ $$T_1 = 100{}^\circ\mathrm{C}$$ $$T_2 = 0{}^\circ\mathrm{C}$$ $$m = 4.8\,\mathrm{kg}$$ $$L = 33600\,\frac{\mathrm{J}}{\mathrm{kg}}$$ $$t = 1\,\mathrm{hour} = 3600\,\mathrm{s}$$

Fourier's law:

$$q = -k\nabla T \equiv k \frac{T_1 - T_2}{d}.$$

Power of a heater:

$$P = qS = \frac{kS(T_1 - T_2)}{d} = \frac{Lm}{t}$$

Thus,

$$k = \frac{Lmd}{S(T_1 - T_2)} = 448\,\frac{\mathrm{W}}{\mathrm{m \cdot K}}$$

Answer:

$$k = 448\,\frac{\mathrm{W}}{\mathrm{m \cdot K}}$$