A container of negligible heat capacity has in it 456 g of ice at - 25 . 0°C . Heat is supplied to the container at the rate of 1000 J / min . After how long will the ice start to melt , assuming all of the ice has the same temperature ? The specific heat of ice is 2090 J / ( kg K ) and the latent heat of fusion of water is 33 . 5 x 104J / kg .

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adam4262 adam4262 · Answer 1 · 2019-12-12T08:45:06+01:00

Answer:

The ice will start to melt after 25.4 mins.

Explanation:

To melt the ice , heat should be supplied such that two processes take place-

i) Increasing the temperature of ice upto 0.0°C

ii) Phase change from ice to water at constant temperature of 0.0°C

Given that,

Initial temperature [tex]T_{1}[/tex] = -25.0°C

Final temperature [tex]T_{2}[/tex] = 0.0°C

mass of ice , m = 456 g = 0.456 kg

Specific heat of ice , s = 2090 J/(kg K)

Latent heat of fusion , L = 33.5 × 104 J/kg

For process i) , heat that should be supplied is -

[tex]Q_{1} = ms\Delta T[/tex] = [tex]ms(T_{2}-T_{1})[/tex] = 0.456×2090×(0-(-25))

[tex]Q_{1}[/tex] = 23826 J

For process ii) , heat that should be supplied is -

[tex]Q_{2} = mL[/tex] = 0.456×33.5 × 104 = 1588.704 J

∴Total heat required is -

[tex]Q_{1}[/tex] + [tex]Q_{2}[/tex] = 25414.704 J

Rate of heat supply (r) = 1000 J/min

∴ The required time (t) is given by -

[tex]t = \frac{25414.704}{r}[/tex] = [tex]\frac{25414.704}{1000}[/tex] = 25.4 mins

∴ t = 25.4 mins