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(a) How much heat should be provided to ice of 720 g mass, lying at -10^@C to melt it to water at 0^@C ?(b) How much heat should be provided to water at 0^@C to increase its temperature to 100^@C ?(c) How much heat should be given to water at 100^@C to transform it completely into water ? (d) Totally, how much heat should be given to ice of 720 g at -10^@C to convert it completely into vapour ? (C"ice"=2220 "J kg"^(-1) K^(-1), C"water"=4190 "J kg"^(-1) K^(-1) ,LF=333 "kJ"/"kg", LV=2256 "kJ"/"kg") |
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Answer» Solution :(a) The TEMPERATURE of ice will not increase until it melts completely. Thus the heat to be given to ice to carry its temperature from `T_i=-10^@C` to `T= 0^@C` (thereafter it will START to melt) is `Q_1=C_"ice"m(T_f-T_i)` where, `C_"ice"`=Specific heat of ice at `-10^@C` `=2220 "J"/"KgK"` `therefore Q_1=2220xx0.720 xx[0-(-10)]` =15984 J `therefore Q_1`= 15.98 kJ Until the ice melts completely, its temperature does not increase above `0^@ C`. Hence, the heat to be given to ice melt it completely is `Q_2=L_F m=(333 kJ//kg)(0.720 kg)` `thereforeQ_2`=239.8 kJ `therefore` Required total heat =`Q_1+Q_2` =15.98 + 239.8 =255.78 kJ (b)Now to increase the temperature of 0.720 kg water from `T_i = 0^@C` to `T_f = 100^@C`, therequired amount of heat is `Q_3=C_"water" m[T_f-T_i]` =4190 X 0.720 x (100-0) `Q_3`=301680 J `therefore Q_3` = 301.68 kJ (c) The heat to be given to water at `100^@C` to transform it completely into vapour is `Q_4=L_V m=2256xx0.720` =1624.32 kJ (d)The total amount of heat to be given to 720 gof ice at `-10^@C` to transform it completely into vapour is `Q=Q_1+Q_2+Q_3+Q_4` =15.98+239.8+301.68+1624.32 =2181.78 kJ |
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