A 2 cm thick slab of ice is at 0 °C throughout. At t = 0 both surfaces of the ice slab are suddenly raised to a temperature of 10 °C and kept at that value thereafter.Determine the time taken for complete melting of the slab and the energy that has entered the slab in this time.

Question

A 2 cm thick slab of ice is at [latex]0^{\circ} C[/latex] throughout. At t = 0 both surfaces of the ice slab are suddenly raised to a temperature of [latex]10^{\circ} C[/latex] and kept at that value thereafter.Determine the time taken for complete melting of the slab and the energy that has entered the slab in this time.

Accepted Answer

We follow the notation used in the text. The properties of ice and water are taken as below

Ice properties:

Density of ice: [latex] ho_{s}=916.8 kg / m ^{3}[/latex]

Latent heat of melting ice: [latex]h_{sf}=3.336 imes 10^{5} J / kg[/latex]

Water properties:

Density of water: [latex] ho_{s}=1000 kg / m ^{3}[/latex]

Specific heat of water: [latex]C_{p f}=4206 J / kg K[/latex]

Thermal diffusivity of ice: [latex]\alpha_{f}=51.535 imes 10^{-6} m ^{2} / s[/latex]

The ice slab thickness is given to be [latex]\delta=2 cm =0.02 m[/latex]. Superheat parameter is zero in this case since [latex]T_{i}=T_{m}=0^{\circ} C[/latex]. The Stefan solution is appropriate. The Stefan number is calculated as

[latex] ext { Ste }=\frac{C_{p f}\left(T_{0}-T_{m} ight)}{h_{s f}}=\frac{4206(10-0)}{3.336 imes 10^{5}}=0.126[/latex]

Linear interpolation using the values in Table 17.2 gives

Table 17.2 Variation with Stefan number of the phase change front

[latex]\mu[/latex]	[latex] ext { Ste }[/latex]	[latex]\mu[/latex]	[latex] ext { Ste }[/latex]	[latex]\mu[/latex]	[latex] ext { Ste }[/latex]
0.01	0.0002	0.14	0.0397	0.6	0.9205
0.02	0.0008	0.16	0.0521	0.7	1.3727
0.03	0.0018	0.18	0.0662	0.8	1.9956
0.04	0.0032	0.2	0.0822	0.9	2.8576
0.05	0.00501	0.22	0.1	1	4.0602
0.06	0.00722	0.24	0.1197	1.2	8.1721
0.07	0.00983	0.26	0.1415	1.4	16.776
0.08	0.0129	0.28	0.1653	1.6	35.8174
0.09	0.0163	0.3	0.1912	1.8	80.5745
0.1	0.0201	0.4	0.3564	2	192.64
0.12	0.0291	0.5	0.5923

[latex]\mu=0.24+\frac{(0.126-0.1197)}{(0.1415-0.1197)}(0.26-0.24)=0.246[/latex]

The charge time is then obtained using Eq. 17.52 as

[latex]t_{c}=\frac{\delta^{2}}{16 \alpha_{f} \mu^{2}}[/latex] (17.52)

[latex]t_{c}=\frac{0.02^{2}}{16 imes 1.535 imes 10^{-6} imes 0.246^{2}}=269.1 s[/latex]

The energy that has entered the ice slab in this time is given by Eq. 17.54 as

[latex]\frac{E_{c}}{ ho_{f} C_{p f}\left(T_{0}-T_{m} ight)}=\frac{1}{\sqrt{\pi}} \cdot \frac{\delta}{\mu \operatorname{erf} \mu}[/latex] (17.54)

[latex]E_{c}=\frac{ ho_{f} C_{p f}\left(T_{0}-T_{m} ight)}{\sqrt{\pi}} \cdot \frac{\delta}{\mu \operatorname{erf} \mu}=\frac{1000 imes 4206 imes(10-0)}{\sqrt{\pi}} imes \frac{0.02}{0.246 imes \operatorname{erf} 0.246}= 7.148 × 10^6 J/m^2[/latex]

Question 17.7: A 2 cm thick slab of ice is at 0 °C throughout. At t = 0 bot...

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