One end of a steel rod (K = 46 J s^{−1} m^{−1}°C^{−1}) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm^{2}. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 10^{5} J kg^{−1}.

#### Solution

Given:

Thermal conductivity of the rod, *K* = 46 J s^{–1} m^{–1} °C^{–1}

Length of the rod, *l* = 1 m

Area of the cross-section of the rod, *A* =0.04 cm^{2}

= 0.04 × 10^{−4} m^{2}

= 4 × 10^{−6} m^{2}

Rate of transfer of heat is given by

`(ΔQ)/(Δt) = (ΔT)/ (l/(KA)`

`(ΔQ)/(Δt) = ((T_1 - T_2 ) KA ) / l`

`(ΔQ)/(Δt) =( (100-0 )/1) xx 46 xx 4 xx 10^-6 m^2`

`(DeltaQ)/(Deltat) = 184xx10^-4 J//s`

Also , ΔQ = mL_{ f}

`therefore (m/t)L_f = 184 xx 10^-4`

`⇒ ( m/t) xx 3.36 xx 10^5 = 184 xx 10^`

`rArr m=(184xx10^-4)/(3.36xx10^5)xxt`

`rArr m = 5.5xx10^-9xx1 kg//s`

`⇒ m = 5.5 xx 10^-8 xx 10^3 g/s`

`⇒ m = 5.5 xx 10^-5 g/s`