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Current Question (ID: 10802)

Question:

$\text{The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are } T_2 \text{ and } T_1 (T_2 > T_1). \text{ The rate of heat transfer through the slab, in a steady state is } \left(\frac{A(T_2-T_1)K}{x}\right)f, \text{ with f which equals to:}$

Options:

1. $1$
2. $\frac{1}{2}$
3. $\frac{2}{3}$
4. $\frac{1}{3}$ (Correct)

Solution:

\text{The problem describes a composite slab made of two materials in series.} \text{We can solve this using the concept of thermal resistance.} \text{The formula for thermal resistance is } R = \frac{L}{kA} \text{1. Resistance of the first slab: } R_1 = \frac{x}{KA} \text{2. Resistance of the second slab: } R_2 = \frac{4x}{2KA} = \frac{2x}{KA} \text{3. Total resistance: } R_{\text{total}} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA} \text{The rate of heat transfer (H) is: } H = \frac{T_2 - T_1}{R_{\text{total}}} H = \frac{T_2 - T_1}{\frac{3x}{KA}} = \frac{KA(T_2 - T_1)}{3x} \text{By comparing with } H = f \frac{A(T_2-T_1)K}{x} \text{, we find f = } \frac{1}{3}

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}