Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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 Cm2. Assuming - Physics

Sum

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 cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 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 cm2
= 0.04 × 10−4 m2
= 4 × 10−6 m2

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

Concept: Heat Transfer - Conduction
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 6 Heat Transfer
Q 5 | Page 98