Import Question JSON

Current Question (ID: 10765)

Question:

$\text{A slab of stone with an area } 0.36 \text{ m}^2 \text{ and thickness of } 0.1 \text{ m is exposed on the lower surface to steam at } 100^\circ\text{C. A block of ice at } 0^\circ\text{C rests on the upper surface of the slab. In one hour } 4.8 \text{ kg of ice is melted. The thermal conductivity of the slab will be:}$ $\text{(Given latent heat of fusion of ice } = 3.36 \times 10^5 \text{ JKg}^{-1})$

Options:

1. $1.29 \text{ J/m/s/}^\circ\text{C}$
2. $2.05 \text{ J/m/s/}^\circ\text{C}$
3. $1.02 \text{ J/m/s/}^\circ\text{C}$
4. $1.24 \text{ J/m/s/}^\circ\text{C}$ (Correct)

Solution:

$\text{Using Fourier's law of heat conduction:}$ $\frac{dQ}{dt} = \frac{KA}{L}\left(T_1 - T_2\right)$ $\text{For steady state heat conduction over time t:}$ $Q = \frac{KA}{L}\left(T_1 - T_2\right)t$ $\text{The heat required to melt the ice:}$ $Q = mL_f$ $\text{Equating the two expressions:}$ $\frac{KA}{L}\left(T_1 - T_2\right)t = mL_f$ $\text{Solving for thermal conductivity K:}$ $K = \frac{mL_f(L)}{A(T_1 - T_2)t}$ $\text{Substituting the given values:}$ $K = \frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600} \text{ J/m/s/C}^0$ $= \frac{4.8 \times 3.36 \times 0.36 \times 36}{4.8}$ $= 1.24 \text{ J/m/s/C}^0$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}