Question

The walls of an ice chest are made of 2.00-mm-thick insulation having a thermal conductivity 0.00300 W/m K. The total surface area of the ice chest is 1.20 m2. If 4.00 kg of ice at 0.00°C are placed in the chest and the temperature of the outside surface of the chest is 20.0°C, how long does it take the ice to melt under steady state conditions? The latent heat of fusion of water is 79.6 cal/g 334 kJ/kg

Answer 1

Answer:

618.5 min

Explanation:

rate of flow of heat =

$\frac{KA(T_2-T_1)}{D}$

Where K is thermal conductivity of separating medium , D is thickness of separation, A is surface area and T₂-T₁ is temperature difference.

Substituting the values given in the problem

we get

Rate of flow

= $\frac{0.003\times1.2\times20}{0.002}$

= 36 J m⁻¹ k⁻¹ s⁻¹

Heat energy required to melt ice

= mass x latent heat of fusion

= 4 x 334000 = 1336000 J

Time required

= heat energy required / rate of flow of heat

= $\frac{1336000}{36}$

= 618.5 min