A copper rod (k=400 W m^-1K^-1, length…

A copper rod (k=400 W m^-1K^-1, length 0.50 m) and a steel rod (k=50 W m^-1K^-1, length 0.25 m), both of cross-sectional area 4 cm^2, are joined end…

Numerical answer

A copper rod (

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

, length

0.50\ \mathrm{m}

) and a steel rod (

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

, length

0.25\ \mathrm{m}

), both of cross-sectional area

4\ \mathrm{cm^2}

, are joined end to end. The free end of the copper rod is held at

100^\circ\mathrm{C}

and the free end of the steel rod is embedded in a large block of ice at

0^\circ\mathrm{C}

. The rods are lagged so that heat flows only along them. In steady state, the mass of ice (in grams) melted in

10

minutes is ______. (Latent heat of fusion of ice

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

. Give the answer correct to two decimal places.)

Show answer and solution

Answer: 11.50

In steady state the same heat current flows through both rods, which behave like thermal resistors in series with

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

. Copper:

R_1=\dfrac{0.5}{400A}

, steel:

R_2=\dfrac{0.25}{50A}

, with

A=4\times10^{-4}\ \mathrm{m^2}

. So

R_1+R_2=\dfrac{1}{A}(0.00125+0.005)=\dfrac{0.00625}{4\times10^{-4}}=15.625\ \mathrm{K\,W^{-1}}

. The heat current is

H=\dfrac{100}{15.625}=6.4\ \mathrm{W}

(the junction is at

80^\circ\mathrm{C}

). In

600\ \mathrm{s}

the heat reaching the ice is

Q=6.4\times600=3840\ \mathrm{J}

. Since all of it melts ice at

0^\circ\mathrm{C}

m=\dfrac{Q}{L_f}=\dfrac{3840}{334000}=0.011497\ \mathrm{kg}\approx11.50\ \mathrm{g}

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