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Current Question (ID: 10757)

Question:

$\text{Two slabs } A \text{ and } B \text{ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab } A \text{ is twice that of } B. \text{ The coefficient of thermal conductivity of slab } A \text{ is twice that of } B. \text{ The first surface of slab } A \text{ is maintained at } 100^\circ\text{C}, \text{ while the second surface of slab } B \text{ is maintained at } 25^\circ\text{C}. \text{ The temperature at the common surface will be:}$

Options:

1. $62.5^\circ\text{C}$ (Correct)
2. $45^\circ\text{C}$
3. $55^\circ\text{C}$
4. $85^\circ\text{C}$

Solution:

$\text{Hint: } H = \frac{KA\Delta T}{L}$ $\text{Step: Find the temperature of the common surface.}$ $\text{Let the common temperature of the surface is } T_s.$ $\text{Thus in the steady state, the rate of heat flowing across a cross-section of the slab is given by;}$ $H = \frac{KA\Delta T}{L}$ $\text{The thermal resistance of the slab is given by;}$ $R_{th} = \frac{L}{KA}$ $\text{The thermal resistance of both slabs is the same i.e., } R_{A_{th}} = R_{B_{th}}$ $\text{The thermal current at the steady state condition is equal i.e.,}$ $H_A = H_B$ $T_i - T_s = T_s - T_f$ $\Rightarrow 100 - T_s = T_s - 25$ $\Rightarrow 2T_s = 125$ $\Rightarrow T_s = \frac{125}{2} = 62.5^\circ\text{C}$ $\text{Hence, option (1) is the correct answer.}$

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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
}