2 kg of ice at `-20^(@)C` is mixed with 5kg of water at `20^(@)C`. The water content of the final mixture is (Latent heat of ice `="80 kcal kg"^(-1)`, the specific heat of water `="1 kcal kg"^(-1).^(@)C^(-1)` and specific heat of ice `="0.5 kcal kg"^(-1).^(@)C^(-1)`)

Options

A. thermal conducitivity of the material is `70J//m.^(@)C`
B. thermal conductivity of the material is `60J//m.^(@)C`
C. mass of the ice melted is 312.5 kg
D. mass of the ice melted is 252 kg

Explanation

Let at any instant temperature of water be T, then heat current
`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`