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Water at 50^(@)C is filled in a cubical container of side 1m. The thickness of the walls of the container is 1 mm. the container is surrounded by large amount of ice at 0.^(@)C the temperature of the water becomes 20.^(@)C in 10ln2 seconds. Choose the correct options. find the thermal conductivity of the material of the container and the ice melted in that time. [Given, specific heat of water =1cal/gm degree, latent heat of fusion of ice =80cal//gm, density of water =1gm//cm^(3) heat capacity of the container cong=0] |
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Answer» THERMAL conducitivity of the material is `70J//m.^(@)C` `i=(kA)/(x).(T-0)`.(1) Where `A=6a^(2)=6m^(2)`= thickness `=1mm=10^(-3)m` Rate of heat lost from water `(dQ)/(dt)=+ms(dT)/(dt)`..(2) So, we get from (1) & (2) `-ms(dT)/(dt)=(kAT)/(x)implies-int_(50^(@))^(25^(@))(dT)/(T)=(kA)/(mSx)int_(0)^(10ln2)dt` `impliesln(2)=(kA)/(mSx).10(ln2)` So, `(kA(10))/(mSx)=1` putting values `impliesk=(mSx)/(10A)=((10^(3)kg)(4.2xx10^(3)J//kg.^(@)C)10^(-3)m)/(10x(6m^(3)))=k=70J//m.^(@)C` `implies` total heat transferred with be = total heat `Q=intdQ=int(kA)/(x)Tdt` lost by water `Q=mStriangleT=10^(3)xx4200xx25J=(10^(6)gm)(1" CAL")(25)=m_(ice)L` Givin `m_(ice)=((10^(6)xx25)/(80))gm=(25000)/(80)kg=312.5kg` Mass of ice melted `=312.5kg` |
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