Import Question JSON

Current Question (ID: 19651)

Question:

$\text{The specific heat of water} = 4200 \text{ Jkg}^{-1}\text{K}^{-1}$ $\text{and the latent heat of ice} = 3.4 \times 10^5 \text{ Jkg}^{-1}$. $100 \text{ grams of ice at } 0^\circ \text{C is placed in } 200 \text{ g of water at } 25^\circ\text{C}$. $\text{The amount of ice that will melt as the temperature of the water reaches } 0^\circ\text{C is close to (in grams):}$

Options:

1. 61.7
2. 69.3
3. 64.6
4. 63.8

Solution:

$\text{Hint: Water will provide heat for ice to melt.}$ $m_w s_w \Delta \theta = m_{\text{ice}} L_{\text{ice}}$ $m_{\text{ice}} = \frac{0.2 \times 4200 \times 25}{3.4 \times 10^5} = 0.0617 \text{ kg} = 61.7 \text{ gm}$ $\text{The remaining ice will remain un-melted.}$ $\text{so correct is (1)}$

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
}