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In an insulated vessel, 250 g of ice at `0^(@)C` is added to 600 g of water at `18^(@)C` (a) What is the final temperature of the system? (b) How much ice remains when the system reaches equilibrium? Useful data: Specific heat capacity of water: 4190 K/K.kg Speicific heat capacity of ice: 2100J/K.kg Latent heat of fusio of ice: `3.34xx10^(5)J//kg` |
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Answer» `Q_("required")=mL` `=(250xx10^(-3))xx3.34xx10^(5)` `=8.35xx10^(4)J`. Heat available from water is `Q_("available")=MCDeltaT=600xx10^(-2)xx4190xx1187` `=4.525xx10^(4)J`. (a) As heast available is not sufficient to melt ice, final temperature is zero. (b) the amount of ice that will melt to bring down the temperature of water from `18^(@)C` to `0^(@)C` is calculated as shown. `Q_("available")=m_(ice)xxL_(ice)` `4.525xx10^(4)=mxx3.34xx10^(5)` `impliesm_(ice)=135xx10^(-1)kg~~135g`. ltBrgt So, ice remaining=250-135 =115g |
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