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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).

Answer» <html><body><p></p>Solution :We have <br/> Mass of the ice, `m = 3` <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> <br/> specific heat <a href="https://interviewquestions.tuteehub.com/tag/capacity-908575" style="font-weight:bold;" target="_blank" title="Click to know more about CAPACITY">CAPACITY</a> of ice, `s_("ice")` <br/> ` = 2100 J kg^(-1) K^(-1)` <br/> specific heat capacity of water , `s_("water")` <br/> ` = 4186 J kg^(-1) K^(-1)` <br/> latent heat of fusion of ice , `L_("f ice ") ` <br/> ` = 3.35 xx 10^(5) J kg^(-1)` <br/> latent heat of steam , `L_("ateam") ` <br/> ` = 2.256 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) J kg^(-1)` <br/> Now, Q = heat required to convert 3 kg of ice at ` - 12^(@) C` to steam at `100 ^(@) C`, <br/> `Q_(1) ` = heat required to convert ice at ` - 12^(@) C` to ice at `0^(@) C` . <br/> `= m s_("ice") Delta T_(1) = (3 kg) (2100 J kg^(-1). K^(-1)) [0 - (-12)]^(@) C = 75600 J` <br/> `Q_(2)` = heat required to <a href="https://interviewquestions.tuteehub.com/tag/melt-1093077" style="font-weight:bold;" target="_blank" title="Click to know more about MELT">MELT</a> ice at `0^(@) C` to water at `0^(@) C` <br/> ` = m L_("f ice ") = (3 kg) (3.35 xx 10^(5) J kg^(-1))` <br/> ` = 1005000 J` <br/> ` Q_(3)` = heat required to convert water at `0^(@) C` to water at `100 ^(@) C`. <br/> ` ms_(w) Delta T_(2) = (3kg) (4186 J kg^(-1) K^(-1)) (100^(@) C)` <br/> ` = 1255800 J` <br/> ` Q_(4)` = heat required to convert water at `100^(@) C` to steam at `100^(@) C`. <br/> ` = m_("steem") = (3 kg) (2.256 xx 10^(6) J kg^(-1))` <br/> ` = 6768000 J ` <br/> So, Q = `Q_(1) + Q_(2) + Q_(3) + Q_(4)` <br/> ` = 75600 J + 100500 J + 1255800 J + 6768000 J` <br/> ` = 9.1 xx 10^(6) J`</body></html>


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