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Current Question (ID: 19658)

Question:

$\text{An ice cube of dimensions } 60 \text{ cm} \times 50 \text{ cm} \times 20 \text{ cm is placed in an insulation box of wall thickness } 1 \text{ cm. The box keeping the ice cube at } 0^\circ\text{C of temperature is brought to a room of temperature } 40^\circ\text{C. The rate of melting of ice is approximately:}$ $\text{(Latent heat of fusion of ice is } 3.4 \times 10^5 \text{ J kg}^{-1} \text{ and thermal conducting of insulation wall is } 0.05 \text{ Wm}^{-1} \text{ } ^\circ\text{C}^{-1})$

Options:

1. $61 \times 10^{-3} \text{ kg s}^{-1}$
2. $61 \times 10^{-5} \text{ kg s}^{-1}$
3. $208 \text{ kg s}^{-1}$
4. $30 \times 10^{-5} \text{ kg s}^{-1}$

Solution:

$\text{The rate of heat transfer through the insulation is given by } Q = \frac{kA\Delta T}{d}.$ $\text{Where } k = 0.05 \text{ Wm}^{-1} \text{ } ^\circ\text{C}^{-1}, A = 0.6 \times 0.5 \text{ m}^2, \Delta T = 40^\circ\text{C}, d = 0.01 \text{ m}.$ $Q = \frac{0.05 \times 0.6 \times 0.5 \times 40}{0.01} = 60 \text{ W}.$ $\text{The rate of melting of ice } = \frac{Q}{L} = \frac{60}{3.4 \times 10^5} = 61 \times 10^{-5} \text{ kg s}^{-1}.$

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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
}