Two rods of the same cross-sectional…

Two rods of the same cross-sectional area 2.0 cm^2 are joined end to end and lagged on their sides. The first is 0.50 m long with thermal…

Numerical answer

Two rods of the same cross-sectional area

2.0\ \mathrm{cm^2}

are joined end to end and lagged on their sides. The first is

0.50\ \mathrm{m}

long with thermal conductivity

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

; the second is

0.25\ \mathrm{m}

long with conductivity

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

. The free end of the first rod is held at

100^\circ\mathrm{C}

by steam and the free end of the second rod is in contact with ice at

0^\circ\mathrm{C}

. After steady state is reached, find the mass of ice, in grams, that melts in

10

minutes. The latent heat of fusion of ice is

334\ \mathrm{kJ\,kg^{-1}}

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 4.75

In steady state the same heat current flows through both rods, so the thermal resistances

R=L/(kA)

add in series.

R_1=\dfrac{0.50}{380\times2.0\times10^{-4}}=6.579\ \mathrm{K\,W^{-1}}

and

R_2=\dfrac{0.25}{40\times2.0\times10^{-4}}=31.25\ \mathrm{K\,W^{-1}}

, giving

R=37.83\ \mathrm{K\,W^{-1}}

. The heat current is

P=\dfrac{\Delta T}{R}=\dfrac{100}{37.83}=2.643\ \mathrm{W}

. In

600\ \mathrm{s}

the heat reaching the ice is

Q=Pt=1586\ \mathrm{J}

, which melts

m=Q/L_f=\dfrac{1586}{3.34\times10^{5}}=4.75\times10^{-3}\ \mathrm{kg}=4.75\ \mathrm{g}

. The junction temperature settles at

100-PR_1\approx82.6^\circ\mathrm{C}

, so most of the temperature drop lies across the poorly conducting second rod, which therefore controls the rate. The ice melts at the rate

P/L_f\approx7.9\times10^{-6}\ \mathrm{kg\,s^{-1}}

, so

600\ \mathrm{s}

gives

4.75\ \mathrm{g}

, in agreement with the value computed above.

Original question written in the JEE Advanced 2018 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake