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Current Question (ID: 10769)

Question:

$\text{One end of a copper rod of uniform cross-section and of length 3.1 m is kept in contact with ice, and the other end with water at } 100°\text{C. At what point along its length should a temperature of } 200°\text{C } \text{be maintained so that in steady-state, the mass of ice melting be equal to that of the steam produced in the same interval time? (Assume that the whole system is insulated from the surroundings. Latent heat of fusion of ice and vaporisation of water are 80 cal/gm and 540 cal/gm respectively)}$

Options:

1. $21.3 \text{ cm from } 100°\text{C end}$ (Correct)
2. $40 \text{ cm from } 0°\text{C end}$
3. $125 \text{ cm from } 100°\text{C end}$
4. $125 \text{ cm from } 0°\text{C end}$

Solution:

\text{(a) Rate of flow of heat is given by } \frac{dQ}{dt} = \frac{\Delta\theta}{l/(KA)} \text{ also } \frac{dQ}{dt} = L \frac{dm}{dt} \text{ (where L = Latent heat)} \Rightarrow \frac{dm}{dt} = \frac{KA}{l} \left(\frac{\Delta\theta}{L}\right) \text{Let the desired point is at a distance x from water at } 100\text{\textdegree C} \therefore \text{ Rate of ice melting = Rate at which steam is being produced} \Rightarrow \left(\frac{dm}{dt}\right)_{\text{Steam}} = \left(\frac{dm}{dt}\right)_{\text{Ice}} \Rightarrow \left(\frac{\Delta\theta}{Ll}\right)_{\text{Steam}} = \left(\frac{\Delta\theta}{Ll}\right)_{\text{Ice}} \Rightarrow \frac{(200-100)}{540 \times x} = \frac{80}{540 \times (3.1-x)} \Rightarrow x = 0.4 \text{ m} = 21.3 \text{ cm}

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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
}