Answer:
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, [tex]L_f[/tex] = 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 = [tex]L_f[/tex] × [tex]m_{melted \ ice}[/tex] =3.35 × 10⁵ J/kg × [tex]m_{melted \ ice}[/tex]
Therefore, from Q = [tex]L_f[/tex] × [tex]m_{melted \ ice}[/tex], we have;
Q = 104976 J = [tex]L_f[/tex] × [tex]m_{melted \ ice}[/tex] = 3.35 × 10⁵ J/kg × [tex]m_{melted \ ice}[/tex]
104976 J = 3.35 × 10⁵ J/kg × [tex]m_{melted \ ice}[/tex]
[tex]m_{melted \ ice}[/tex] = 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg
The mass of melted ice, [tex]m_{melted \ ice}[/tex] ≈ 0.31336 kg
∴ The amount of solid ice remaining after 6 hours, [tex]m_{ice}[/tex] = m - [tex]m_{melted \ ice}[/tex]
Which gives;
[tex]m_{ice}[/tex] = m - [tex]m_{melted \ ice}[/tex] = 4.0 kg - 0.31336 kg ≈ 3.68664 kg
The amount of solid ice remaining after 6 hours, [tex]m_{ice}[/tex] ≈ 3.68664 kg.
