JEE Advanced 2019 Paper 2 · Physics · Calorimetry and Heat Transfer · medium

A copper rod and a steel rod, each 0.50 m long with the same cross-section 4.0 cm^2, are joined end to end. The free end of the copper rod is kept in…

Numerical answer

A copper rod and a steel rod, each

0.50\ \mathrm{m}

long with the same cross-section

4.0\ \mathrm{cm^2}

, are joined end to end. The free end of the copper rod is kept in steam at

100^\circ\mathrm{C}

and the free end of the steel rod is in contact with melting ice at

0^\circ\mathrm{C}

. The thermal conductivities are

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

(copper) and

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

(steel); the sides are perfectly insulated and steady state is reached. The latent heat of fusion of ice is

336\ \mathrm{J\,g^{-1}}

. The mass of ice melted in

10

minutes, in grams, is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 6.35

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

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

. For copper,

R_1=\dfrac{0.5}{400\times4\times10^{-4}}=3.125\ \mathrm{K\,W^{-1}}

; for steel,

R_2=\dfrac{0.5}{50\times4\times10^{-4}}=25\ \mathrm{K\,W^{-1}}

. The total is

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

, so the heat current is

\dot Q=\dfrac{100}{28.125}=3.556

W (the interface settles near

88.9^\circ\mathrm{C}

). In

600

s the heat reaching the ice is

2133

J, which melts

m=\dfrac{2133}{336}=6.35

g. Using only the steel rod's resistance, a common error, would overstate the rate. Equivalent resistances add exactly as in series electrical circuits, with temperature playing the role of potential and heat current that of electric current.

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