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Current Question (ID: 10738)

Question:
$\text{5 g of water at 30°C and 5 g of ice at -20°C are mixed together in a calorimeter. The water equivalent of the calorimeter is negligible, and the specific heat and latent heat of ice are 0.5 cal/g°C and 80 cal/g, respectively. The final temperature of the mixture is:}$
Options:
  • 1. $0°\text{C}$ (Correct)
  • 2. $-8°\text{C}$
  • 3. $-4°\text{C}$
  • 4. $2°\text{C}$
Solution:
$\text{Hint: Assume final temp. of mixture as 0°C and check}$ $\text{Step 1: Find heat required to convert ice to water to 0°C}$ $-20°\text{C} \xrightarrow{Q_1} 0°\text{C} \xrightarrow{Q_2} 0°\text{C}$ $\text{ice} \quad \text{water}$ $Q_1 = 50 \text{ cal}$ $Q_{\text{ice}} = Q_1 + Q_2$ $= mc \Delta T + mL_f$ $= 5 \times 0.5 \times 20 + 5 \times 80$ $= 450 \text{ cal}$ $\text{Step 2: Find heat released by water}$ $30°\text{C} \xrightarrow{} 0°\text{C}$ $\text{water}$ $Q_w = mc \Delta T$ $= 5 \times 1 \times 30$ $= 150 \text{ cal}$ $\text{Step 3: Check if heat released by water is sufficient to melt all ice}$ $Q_w > Q_{\text{ice}}$ $150 < 450$ $\text{So, the final temperature of the mixture is 0°C.}$

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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
}