Question

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​

Answer 1

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

Step-by-step 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,
`L_f` = 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 =
`L_f` ×
`m_(melted \ ice)` =3.35 × 10⁵ J/kg ×
`m_(melted \ ice)`

Therefore, from Q =
`L_f` ×
`m_(melted \ ice)`, we have;

Q = 104976 J =
`L_f` ×
`m_(melted \ ice)` = 3.35 × 10⁵ J/kg ×
`m_(melted \ ice)`

104976 J = 3.35 × 10⁵ J/kg ×
`m_(melted \ ice)`

`m_(melted \ ice)` = 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg

The mass of melted ice,
`m_(melted \ ice)` ≈ 0.31336 kg

∴ The amount of solid ice remaining after 6 hours,
`m_(ice)` = m -
`m_(melted \ ice)`

Which gives;

`m_(ice)` = m -
`m_(melted \ ice)` = 4.0 kg - 0.31336 kg ≈ 3.68664 kg

The amount of solid ice remaining after 6 hours,
`m_(ice)` ≈ 3.68664 kg.