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`        1 kg ice at -10 degree celsiusis mixed with 1kg water at 100 degree celsius .then find temperature at equilibrium and mixture content/`
9 months ago

Arun
22572 Points
```							Dear Diksha The idle warmth of dissolving of water is 334 kJ/kg So 1 kg of water at 0 deg. C contains 334 kJ more warmth vitality than 1 kg of ice at 0 deg. C. The particular warmth (warm limit) of water varies with temperature yet not by much. In the temperature extend from 0 to 100 deg. C, it is approx. 4.2 kJ/(kg K). So 1 kg of water at 100 deg. C contains 4.2 x 100 = 420 kJ more warmth vitality than 1 kg of water at 0 deg. C. The inert warmth of dissipation of water is 2257 kJ/kg For 1 kg of steam at 100 deg. C contains 2257 kJ more warmth vitality than 1 kg of water at 100 deg. C. Therefore 1 kg of steam at 100 deg. C has 334 + 420 + 2257 = 3011 kJ more warmth vitality than 1 kg of ice at 0 deg. C. 1 kg of ice and 1 kg of steam = 2 kg of blend. 3011 kJ in 2 kg = 3011/2 = 1505.5 kJ/kg. We require the temperature at which 1 kg of water (or ice or steam) has 1505.5 kJ more warmth vitality than 1 kg of ice at 0 deg. C. 1505.5 - 334 = 1171.5 So it is likewise the temperature at which 1 kg of water (or steam) has 1171.5 kJ more warmth vitality than 1 kg of water at 0 deg. C. 1171.5/4.2 = 278.9 278.9 greater than 100 Hence the temperature of the mix is surely 100 deg. C. You require more vitality to bubble water than to get it from ice to breaking point! RegardsArun (askIITians forum expert)
```
9 months ago
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