A conducting rod of length 1 m has area of cross-section 10⁻³ m². One end is immersed in boiling water (100) and the other end in Ice (0). If coefficient of thermal conductivity of rod is 96 cal/s m and latent heat for ice is 8 × 10⁴ cal/kg then the amount of ice which will melt in one minute is (a) 5.4 × 10⁻³ kg (b) 7.2 × 10⁻³ kg (c) 1.8 × 10⁻³ kg (d) 3.6 × 10⁻³ kg

Answer: (b). Rate of heat flow H = (KAΔ T)/(L) = 96×10⁻³×1001 = 9.6 cal/s Heat in 60 s: Q = 9.6×60 = 576 cal m = Q/Lf = 5768×10⁴ = 7.2×10⁻³ kg So the answer is (b) 7.2×10⁻³ kg.

A conducting rod of length 1 m has area of cross-section 10⁻³ m². One… | MHT CET 2021 (20 Sep, Shift 2)

Q41MHT CET 2021 · 20 SEP · SHIFT 2Difficult

Heat Conduction

A conducting rod of length $1\text{ m}$ has area of cross-section $10^{-3}\text{ m}^2$ . One end is immersed in boiling water $(100^\circ\text{C})$ and the other end in Ice $(0^\circ\text{C})$ . If coefficient of thermal conductivity of $\text{rod}$ is $96\text{ cal/s m}^\circ\text{C}$ and latent heat for ice is $8 \times 10^{4}\text{ cal/kg}$ then the amount of ice which will melt in one minute is

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Answer: (b)

Rate of heat flow $H = \dfrac{KA\Delta T}{L} = \dfrac{96\times10^{-3}\times100}{1} = 9.6$ cal/s
Heat in 60 s: $Q = 9.6\times60 = 576$ cal
$m = \dfrac{Q}{L_f} = \dfrac{576}{8\times10^{4}} = 7.2\times10^{-3}$ kg
So the answer is (b) $7.2\times10^{-3}$ kg.

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Heat Conduction question from MHT CET 2021 (20 Sep, Shift 2)

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