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When 0.15 kg of ice of "0 "^(@)C mixed with 0.30 kg of water at "50 "^(@)C in a container, the resulting temperature is 6.7" "^(@)C. Calculate the heat of fusion of ice. (s_("water")=4186" J Kg"^(-1)K^(-1))

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of <a href="https://interviewquestions.tuteehub.com/tag/ice-1035302" style="font-weight:bold;" target="_blank" title="Click to know more about ICE">ICE</a> `m_(1)=0.15` kg <br/> Temperature of ice `t_(1)=0^(@)C` <br/> Mass of water `m_(2)=0.30` kg <br/> Temperature of water `t_(2)=50^(@)C` <br/> Temperature of mixture of water and ice `t_(3)=6.7^(@)C` <br/> Specific <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> of water `S=4186" Jkg"^(-1)K^(-1)` <br/> Expose latent heat of fusion of ice is `L_(f)` <br/> `rArr` 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> the ice, <br/> `Q_(1)=m_(1)L_(f)` <br/> `:.Q_(1)=0.15L_(f)J". . . (1)"` <br/> `rArr` Heat loss by water, <br/> `Q_(2)=m_(2)S(t_(2)-t_(3))` <br/> `=0.30xx4186(50-6.7)` <br/> `=0.30xx4186xx43.3` <br/> `:.Q_(2)=54376.14J`. . .(2) <br/> Heat required to make temperature of water of ice `6.7^(@)C`, <br/> `Q_(3)=m_(1)S(t_(3)-t_(1))` <br/> `=0.15xx4186(6.7-0)` <br/> `=0.15xx4186xx6.7` <br/> `:.Q_(3)=4206.93J`. . . (3) <br/> {Heat <a href="https://interviewquestions.tuteehub.com/tag/lost-537630" style="font-weight:bold;" target="_blank" title="Click to know more about LOST">LOST</a> by water} = {Heat required to melt the ice} + {Heat required to make temperature of water to ice `t_(3)`} <br/> `Q_(2)=Q_(1)+Q_(3)` <br/> `:.54376.14=0.15L_(f)=4206.96""` [from equation (1), (2) and (3)] <br/> `:.0.15L_(f)=54376.144206.96` <br/> `=50xx169.21` <br/> `:.L_(f)=(50xx169.21)/(0.15)=334461.4" Jkg"^(-1)~~3.34xx10^(5)" Jkg"^(-1)`</body></html>


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