Import Question JSON

Current Question (ID: 8246)

Question:

$\text{The amount of heat needed to raise the temperature of } 60.0 \, \text{g of aluminium from } 35°\text{C to } 55°\text{C would be:}$\n\n$(\text{Molar heat capacity of Al is } 24 \, \text{J mol}^{-1} \text{K}^{-1})$

Options:

1. $1.07 \, \text{J}$
2. $1.07 \, \text{kJ}$
3. $106.7 \, \text{kJ}$
4. $100.7 \, \text{kJ}$

Solution:

\textbf{Hint:} \, q = n \cdot c \cdot \Delta T \textbf{Step 1:} \text{ Write down the general formula for heat in terms of heat capacity} \text{From the expression of heat } (q)\text{,} q = n \cdot c \cdot \Delta T \ldots\ldots\ldots\ldots(\text{I}) \text{Where,} c = \text{molar heat capacity; } n = \text{number of moles and } \Delta T = \text{change in temperature} \textbf{Step 2:} \text{ Substitute the given value in (I)} \text{First, calculate number of moles of Al:} n = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0 \text{ g}}{27 \text{ g/mol}} = \frac{60}{27} \text{ mol} \text{Calculate temperature change:} \Delta T = 55^\circ\text{C} - 35^\circ\text{C} = 20^\circ\text{C} = 20 \text{ K} \text{Now substitute in the formula:} q = \left(\frac{60}{27} \text{ mol}\right)(24 \text{ J mol}^{-1} \text{K}^{-1})(20 \text{ K}) q = \frac{60 \times 24 \times 20}{27} \text{ J} = \frac{28800}{27} \text{ J} = 1066.7 \text{ J} = 1.07 \text{ kJ} \text{Therefore, option (2) is correct.}

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
}