JEE Advanced 2018 Paper 1 · Physics · Calorimetry and Heat Transfer · medium

A copper rod of length 0.40 m and a steel rod of length 0.30 m, each of cross-sectional area 2.0 cm^2, are joined end to end and thermally insulated…

Numerical answer

A copper rod of length

0.40\ \mathrm{m}

and a steel rod of length

0.30\ \mathrm{m}

, each of cross-sectional area

2.0\ \mathrm{cm^2}

, are joined end to end and thermally insulated on their sides. The free end of the copper rod is held at

100^\circ\mathrm{C}

and that of the steel rod is in contact with a large block of melting ice at

0^\circ\mathrm{C}

. The thermal conductivities are

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

for copper and

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

for steel, and the latent heat of fusion of ice is

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

. Find the mass of ice, in grams, melted by the rods in

10

minutes of steady state. Give the answer correct to two decimal places.

Show answer and solution

Answer: 5.10

In steady state the rods behave as thermal resistances in series,

R_{th}=\dfrac{L}{kA}

. For copper

R_1=\dfrac{0.40}{385\times2\times10^{-4}}=5.195\ \mathrm{K\,W^{-1}}

and for steel

R_2=\dfrac{0.30}{50\times2\times10^{-4}}=30.0\ \mathrm{K\,W^{-1}}

. The total is

35.195\ \mathrm{K\,W^{-1}}

, so the heat current is

P=\dfrac{100}{35.195}=2.841\ \mathrm{W}

(the interface is at about

85.2^\circ\mathrm{C}

). In

600\ \mathrm{s}

the heat delivered to the ice is

Q=2.841\times600=1704.7\ \mathrm{J}

. The ice melted is

m=Q/L_f=1704.7/(3.34\times10^5)=5.10\times10^{-3}\ \mathrm{kg}=5.10\ \mathrm{g}

Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake