Question

A slab of ice (100 mmsquare and10 mmthick) is placed in a container with the top surface open to the ambient air(T∞=25°C,h=25 W/m2.K).The other surfaces are properly insulated. Assuming ice-water mixture remains at 0°C, how long will it take to completely melt the ice? Density and latent heat of fusion of ice are 920 kg/m3 and 334 kJ/kgrespectively.​

Answer 1

Answer:

412.456

Explanation:

Q' = h*A*(T2 - T1)

Q' = m'L

(920 * (10^(-3)) * 334 * 10^3) / t = 25 * 0.1 * (298 - 0)

-> t = 412.45 s

Answer 2

9833 seconds

Explanation:

It is given that,

A slab of ice with thickness = 10mm

Size of slab of ice = 100mm

T∞=25°C,h=25 W/$m^{2}*K$

Density of ice = 920 kg/$m^{3}$

Latent heat of fusion of ice = 334 KJ/Kg

Solution:

We need to find how long will it take to completely melt the ice

Latent heat of fusion of ice: Ice has a latent heat of fusion of 33600 J/K. The amount of heat required to melt a unit mass of ice from the solid to the liquid state is known as the latent heat of fusion of ice.

The rate of heat transfer of ice is given by 'q',

⇒ q = $hA$(T∞ - T)

In this equation, h represents the heat transfer coefficient and A represents the ice's top surface. The amount of energy required to melt the ice is pALΔ*$h_{fusion}$ where L is the ice slab's thickness The time has come to melt the ice.

⇒ $t = \frac{ p*AL(delta)*h_{fusion}}{hA(T_{\alpha } - T}$

⇒ $\frac{(920)(0.10)(334000)}{25(25)}$

⇒ 9833 s.

Hence, the time taken by ice to melt completely is 9833 seconds.

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