One end of a copper rod of length 1 m and area of cross - section `400 xx 10^-4 m^2` is maintained at `100^@C`. At the other end of the rod ice is kept at `0^@C`. Neglecting the loss of heat from the surrounding, find the mass of ice melted in 1h. Given, `K_(Cu) = 401 W//m.K and L_f = 3.35 xx 10^5 J//kg.`

Explanation

Thermal resistance of the rod, `R=l/(KA)=(1.0)/(401)(4 xx 10^(-4)=6.23 KW^(-1)`
`therefore` Heat current, h=(Temperature difference)/(Thermal resistance)`
`(100-0)/6.23 = 16W`
Heat transferred in 1 h,
Q= Ht
(`therefore H=Q/t)`
Now, let m be the mass of ice that melts in 1h, then
`m=Q/L` (Q=mL)
`=(57600/(3.35 xx 10^(5)` = 0.172 kg or 172 g