A cube icebox of side 3cm has a thickness of 5.0cm. If 4.0 kg of ice is put in the box estimate the amount of ice remaining after 6hrs. The outside T° is 45°c and coefficient of thermal conductivity of thermacole is 0.01J /s/m/k. The heat of fusion of water is 3.35× 10^5 J/k/hr/kg

The amount of solid ice remaining after 6 hours is approximately 3.68664 kg

Explanation:

The given parameters are;

The side length of the cube box, s = 3(0) cm = 0.3 m

The thickness of the cube box, d = 5.0 cm = 0.05 m  

The mass of ice in the box, m = 4.0 kg

The outside temperature of the cube box, T₁ = 45°C

The temperature of the melting ice inside the box, T₂ = 0°C

The latent heat of fusion of ice, = 3.35 × 10⁵ J/K/hr/kg

The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²

The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹

For thermal equilibrium, we have;

The heat supplied by the surrounding = The heat gained by the ice

The  heat supplied by the surrounding, Q = K·A·ΔT·t/d

Where;

ΔT = T₁ - T₂ =  45° C - 0° C = 45° C

ΔT = 45° C

Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976

∴ The  heat supplied by the surrounding, Q = 104976 J

The heat gained by the ice = × =3.35 × 10⁵ J/kg ×

Therefore, from Q =   × , we have;

Q = 104976 J =   × = 3.35 × 10⁵ J/kg ×

104976 J = 3.35 × 10⁵ J/kg ×

= 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg

The mass of melted ice, ≈ 0.31336 kg

∴ The amount of solid ice remaining after 6 hours, = m -

Which gives;

= m - = 4.0 kg - 0.31336 kg ≈ 3.68664 kg

The amount of solid ice remaining after 6 hours, ≈ 3.68664 kg.

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

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