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Calculate the heat required to convert 3 kg of ice at -12" "^(@)C kept in a calorimeter to steam at 100" "^(@)C at atmospheric pressure. Given specific heat capacity of ice =2100" J kg"^(-1)K^(-1), specific heat capacity of water =4186" J kg"^(-1)K^(-1), latent heat of fusionn of ice =3.35xx10^(5)" J kg"^(-1) and latent heat of steam =2.256xx10^(6)" J kg"^(-1). |
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Answer» Solution :Mass of ice `m=3` kg Specific heat of ice `S_("ice")=2100" Jkg"^(-1)K^(-1)` Specific heat of water `S_(W)=4186" Jkg"^(-1)K^(-1)` Latent heat of fusion of ice `L_(F"(ice)")=3.35xx10^(5)" Jkg"^(-1)` Latent heat of VAPORIZATION of steam `L_("heat")=2.256xx10^(6)" Jkg"^(-1)` `rArrQ_(1)=` heat required to rise temperature of ice from `-12^(@)C` to `0^(@)C` `="mS"_("ice")DeltaT_(1)` `=3xx2100xx[0-(-12)]` `=6300xx12` `:.Q_(1)=756000" J. . . (1)"` `rArr Q_(2)=` heat required to CONVERT ice of `0^(@)C` in to water `=mL_("ice")` `3xx3.35xx10^(5)` `:.Q_(2)=1005000 " J. . .(2)"` `rArrQ_(3)=` heat required to rise temperature of water from `0^(@)C` to `100^(@)C` `=mS_(W)DeltaT_(2)` `=3xx4186xx(100-0)` `:.Q_(3)=1255800" J. . .(3)"` `rArr Q_(4)` = heat required to convert water of `100^(@)C` in to steam `=mL_("steam")` `=3xx2.256xx10^(6)` `:.Q_(4)=" 6768000" J. . .(4)"` `rArrQ=` heat required to convert ice at `-12^(@)C` in to steam `:.Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)` From equation (1), (2), (3) and (4) `Q=75600+1005000+1255800+6768000=9104400" J"` `:.Q~~9.1xx10^(6)J` |
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