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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)) |
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Answer» Solution :MASS of ICE `m_(1)=0.15` kg Temperature of ice `t_(1)=0^(@)C` Mass of water `m_(2)=0.30` kg Temperature of water `t_(2)=50^(@)C` Temperature of mixture of water and ice `t_(3)=6.7^(@)C` Specific HEAT of water `S=4186" Jkg"^(-1)K^(-1)` Expose latent heat of fusion of ice is `L_(f)` `rArr` Heat required to MELT the ice, `Q_(1)=m_(1)L_(f)` `:.Q_(1)=0.15L_(f)J". . . (1)"` `rArr` Heat loss by water, `Q_(2)=m_(2)S(t_(2)-t_(3))` `=0.30xx4186(50-6.7)` `=0.30xx4186xx43.3` `:.Q_(2)=54376.14J`. . .(2) Heat required to make temperature of water of ice `6.7^(@)C`, `Q_(3)=m_(1)S(t_(3)-t_(1))` `=0.15xx4186(6.7-0)` `=0.15xx4186xx6.7` `:.Q_(3)=4206.93J`. . . (3) {Heat LOST by water} = {Heat required to melt the ice} + {Heat required to make temperature of water to ice `t_(3)`} `Q_(2)=Q_(1)+Q_(3)` `:.54376.14=0.15L_(f)=4206.96""` [from equation (1), (2) and (3)] `:.0.15L_(f)=54376.144206.96` `=50xx169.21` `:.L_(f)=(50xx169.21)/(0.15)=334461.4" Jkg"^(-1)~~3.34xx10^(5)" Jkg"^(-1)` |
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