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Calculate the heat required to convert 3 kg of ice at -12 ^(@) Ckept 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 fusion of ice= 3.35 xx 10^(5) J kg^(-1) and latent heat of steam= 2.256 xx 10^(8) J kg^(-1). |
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Answer» Solution :We have Mass of the ice, `m = 3` KG specific heat CAPACITY of ice, `s_("ice")` ` = 2100 J kg^(-1) K^(-1)` specific heat capacity of water , `s_("water")` ` = 4186 J kg^(-1) K^(-1)` latent heat of fusion of ice , `L_("f ice ") ` ` = 3.35 xx 10^(5) J kg^(-1)` latent heat of steam , `L_("ateam") ` ` = 2.256 xx 10^(6) J kg^(-1)` Now, Q = heat required to convert 3 kg of ice at ` - 12^(@) C` to steam at `100 ^(@) C`, `Q_(1) ` = heat required to convert ice at ` - 12^(@) C` to ice at `0^(@) C` . `= m s_("ice") Delta T_(1) = (3 kg) (2100 J kg^(-1). K^(-1)) [0 - (-12)]^(@) C = 75600 J` `Q_(2)` = heat required to MELT ice at `0^(@) C` to water at `0^(@) C` ` = m L_("f ice ") = (3 kg) (3.35 xx 10^(5) J kg^(-1))` ` = 1005000 J` ` Q_(3)` = heat required to convert water at `0^(@) C` to water at `100 ^(@) C`. ` ms_(w) Delta T_(2) = (3kg) (4186 J kg^(-1) K^(-1)) (100^(@) C)` ` = 1255800 J` ` Q_(4)` = heat required to convert water at `100^(@) C` to steam at `100^(@) C`. ` = m_("steem") = (3 kg) (2.256 xx 10^(6) J kg^(-1))` ` = 6768000 J ` So, Q = `Q_(1) + Q_(2) + Q_(3) + Q_(4)` ` = 75600 J + 100500 J + 1255800 J + 6768000 J` ` = 9.1 xx 10^(6) J` |
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