A glass contains a mixture of water and ice cubes, with a total mass of 0 . 2 kg

Question

A glass contains a mixture of water and ice cubes, with a total mass of 0.2 kg and at a temperature of

0 C. The mass of ice cubes is 1/3 of the mass of the mixture in the glass. As a result of heat transfer

from the surroundings at 20 Celsius degree to the mixture, the ice has molten and the temperature of the water

increased to 20 Celsius degree. Assume the latent heat of melting ice equal to 333 kJ kg?1.

Question:

Calculate the entropy change of the substance in the glass and of the surroundings.

Explanation / Answer

Solution:

let Mw = mass of water and Mi = mass of ice; Mi = (1/3) (Mw + Mi)

Mw + Mi = 0.2 and Mi = 0.067 Kg also Mw = 0.133 Kg

Total heat taken by the mixture from the surronding = Latent heat of exchange by 0.067 Kg of ice to melt in 0 degree celsius of water + Sensible heat transfer by 0.2 Kg of water at 0 degree celsius upto 20 degree Celsius

= Mi * Latent heat + (Mw + Mi) * 4.18 *(20 - 0).. susbtituting all in this equation we get

Total heat taken by the mixture from the surronding = 39.031 KJ

Entropy change of the ice: (0.067*333)/(0+273) = 0.082 KJ / K

Entropy change of the water, from zero to 20 degree celsius =

= change of entropy of the mix = 0.2* 4.18* ln (293/273) = 0.06 KJ / K

Therefore;

entropy change of the mixture: 0.082+0.06 = 0.142 KJ / K

Entropy change of the surronding = -Qtotal / Tsurronding = -39.031/(20+273) = -0.133 KJ / K

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