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Difficulty: HardModes of Heat Transfer (Conduction, Convection, and Radiation)

A solid cylindrical metal rod of length 0.50 m0.50\text{ m} and cross-sectional area 4.0×10−4 m24.0 \times 10^{-4}\text{ m}^2 is perfectly insulated along its lateral surface. One end of the rod is maintained at 100∘C100^\circ\text{C} in boiling water, while the other end is in contact with an ice block at 0∘C0^\circ\text{C}. If 0.072 kg0.072\text{ kg} of ice melts in 10 minutes10\text{ minutes} due to thermal energy conducted through the rod, what is the thermal conductivity of the metal? (Take the specific latent heat of fusion of ice as 3.36×105 J⋅kg−13.36 \times 10^5\text{ J}\cdot\text{kg}^{-1})

  1. A
    630 W⋅m−1⋅K−1630\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}
  2. 504 W⋅m−1⋅K−1504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}Answer
  3. C
    7000 W⋅m−1⋅K−17000\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}
  4. D
    3.23 W⋅m−1⋅K−13.23\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

Answer

The thermal conductivity of the metal rod is 504 W⋅m−1⋅K−1504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}.
The correct answer is derived by first finding the total heat absorbed during the phase change of ice using Q=mL=0.072×3.36×105=24,192 JQ = m L = 0.072 \times 3.36 \times 10^5 = 24,192\text{ J}. Dividing by the time in seconds (600 s600\text{ s}) gives a heat flow rate of 40.32 W40.32\text{ W}. Substituting this into the conduction formula Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d} gives 40.32=k(4.0×10−4)(100)0.50=0.08k40.32 = \frac{k (4.0 \times 10^{-4})(100)}{0.50} = 0.08 k, yielding k=504 W⋅m−1⋅K−1k = 504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}.

Step-by-Step Solution

1
Calculate the total heat energy QQ required to melt 0.072 kg0.072\text{ kg} of ice at 0∘C0^\circ\text{C}.
Q=mL=0.072 kg×3.36×105 J⋅kg−1=24,192 JQ = m L = 0.072\text{ kg} \times 3.36 \times 10^5\text{ J}\cdot\text{kg}^{-1} = 24,192\text{ J}
During melting at constant temperature, heat transfer is governed by the latent heat of fusion formula.
2
Convert the elapsed time into seconds and calculate the rate of heat transfer Qt\frac{Q}{t}.
t=10 min=600 st = 10\text{ min} = 600\text{ s}; Qt=24,192 J600 s=40.32 W\frac{Q}{t} = \frac{24,192\text{ J}}{600\text{ s}} = 40.32\text{ W}
Thermal conductivity formulas require rate of heat transfer in Joules per second (Watts).
3
Apply Fourier's law of thermal conduction Qt=kA(T1−T2)d\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d} to solve for thermal conductivity kk.
40.32=k×(4.0×10−4)×(100−0)0.50  ⟹  40.32=0.08k  ⟹  k=504 W⋅m−1⋅K−140.32 = \frac{k \times (4.0 \times 10^{-4}) \times (100 - 0)}{0.50} \implies 40.32 = 0.08 k \implies k = 504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}
Rearranging the steady-state thermal conduction equation yields k=(Q/t)⋅dA⋅ΔTk = \frac{(Q/t) \cdot d}{A \cdot \Delta T}.

Key Concept

Thermal Conduction Rate and Latent Heat of Fusion
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Topics

Modes of Heat Transfer (Conduction, Convection, and Radiation)

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