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Current Question (ID: 10745)

Question:

$\text{A geyser heats water flowing at a rate of 3.0 litres per minute from 27°C to 77°C. If the geyser operates on a gas burner and its heat of combustion is } 4.0 \times 10^4 \text{ J/g, then what is the rate of combustion of fuel nearly?}$

Options:

1. $24 \text{ g/min}$
2. $12 \text{ g/min}$
3. $32 \text{ g/min}$
4. $16 \text{ g/min}$ (Correct)

Solution:

$\text{Hint: Heat produced by fuel = Heat gained by the water}$ $\text{Step 1: Find mass flow rate for water}$ $\frac{m}{t} = \frac{V}{t} \times \rho = \frac{3l}{min} \times 1000 \frac{g}{l} = 3000 \text{ g/min}$ $\text{Step 2: Find the rate of heating of water}$ $\frac{Q_1}{t} = \frac{m}{t} \times c \times \Delta T$ $= 3000 \times 4.2 \times 50 \text{ J/min}$ $\text{where } c = 4.2 \text{ J/g°C (specific heat of water) and } \Delta T = 77 - 27 = 50°C$ $\text{Step 3: Find the rate of heat supplied by fuel}$ $\frac{Q_2}{t} = \frac{m}{t} \times 4 \times 10^4 \frac{g}{min} \times \frac{J}{g}$ $\text{Step 4: Apply energy conservation}$ $\text{Rate of heat supplied by fuel = Rate of heating of water}$ $\left(\frac{m}{t}\right) \times 4 \times 10^4 = 3000 \times 4.2 \times 50$ $\frac{m}{t} = \frac{3000 \times 4.2 \times 50}{4 \times 10^4} = 15.75 \text{ g/min}$ $\approx 16 \text{ g/min}$

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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
}