Import Question JSON

Current Question (ID: 10747)

Question:

$\text{From the given heat and temperature graph of a substance, choose the correct statement:}$

Options:

1. $\text{The latent heat of fusion is greater than the latent heat of vaporization.}$
2. $\text{The latent heat of vaporization is greater than the latent heat of fusion.}$
3. $\text{The specific heat in a solid state is more than the specific heat in a liquid state.}$ (Correct)
4. $\text{The specific heat in a liquid state is more than the specific heat in a solid state.}$

Solution:

\text{Hint: During phase change, temperature does not change.} \text{Step 1: Identify phase change parts in the graph} \text{From the graph: Solid phase (0 to H), fusion (H to 2H), liquid phase (2H to 2.5H), vaporization (2.5H to 3.5H)} \text{Step 2: Calculate latent heats} \text{For fusion: } Q = mL_f \text{, so } H = mL_f \text{ which gives } L_f = \frac{H}{m} \text{For vaporization: } Q = mL_v \text{, so } H = mL_v \text{ which gives } L_v = \frac{H}{m} \text{From the graph, heat required for vaporization (2.5H to 3.5H) = H} \text{Heat required for fusion (H to 2H) = H} \text{Therefore: } L_f = L_v = \frac{H}{m} \text{Step 3: Calculate specific heats} \text{For solid: } Q = mc_s\Delta T \text{, slope } = \frac{\Delta Q}{\Delta T} = mc_s \text{From graph: } c_s = \frac{H}{m\theta} \text{For liquid: } c_l = \frac{H}{2m\theta} \text{Since } \frac{H}{2m\theta} < \frac{H}{m\theta} \text{, we have } c_l < c_s \text{Therefore, the specific heat in a solid state is more than the specific heat in a liquid state.}

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
}