Two rods of equal cross-section… | JEE Advanced 2025 P2

Two rods of equal cross-section 1.0×10^-4 m^2 are joined end to end. Rod 1 has length 0.40 m and thermal conductivity 200 W m^-1K^-1; rod 2 has…

Numerical answer

Two rods of equal cross-section

1.0\times10^{-4}\ \mathrm{m^2}

are joined end to end. Rod 1 has length

0.40\ \mathrm{m}

and thermal conductivity

200\ \mathrm{W\,m^{-1}K^{-1}}

; rod 2 has length

0.60\ \mathrm{m}

and conductivity

100\ \mathrm{W\,m^{-1}K^{-1}}

. The free end of rod 1 is held at

100^\circ\mathrm{C}

and the free end of rod 2 is in contact with a large block of ice at

0^\circ\mathrm{C}

. The sides are insulated and the steady state has been reached. Latent heat of fusion of ice is

3.36\times10^{5}\ \mathrm{J\,kg^{-1}}

. How long, in minutes, does it take to melt

10\ \mathrm{g}

of ice? Give the answer correct to two decimal places.

Show answer and solution

Answer: 44.80

In steady state the heat current is the same through both rods and the thermal resistances add in series.

R_1=\dfrac{L_1}{k_1A}=\dfrac{0.40}{200\times10^{-4}}=20\ \mathrm{K\,W^{-1}}

and

R_2=\dfrac{0.60}{100\times10^{-4}}=60\ \mathrm{K\,W^{-1}}

. Total

R=80\ \mathrm{K\,W^{-1}}

, so

H=\dfrac{100}{80}=1.25\ \mathrm{W}

(junction temperature is

100-1.25\times20=75^\circ\mathrm{C}

). Heat needed to melt

10\ \mathrm{g}

Q=mL_f=0.010\times3.36\times10^5=3360\ \mathrm{J}

. Time

t=Q/H=3360/1.25=2688\ \mathrm{s}=44.80\ \mathrm{min}

. The ice stays at

0^\circ\mathrm{C}

so the end temperature of rod 2 is fixed.

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