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Current Question (ID: 19633)

Question:

$\text{In an experiment to verify Newton's law of cooling, a graph is plotted}$ $\text{between the temperature difference } (\Delta T) \text{ of the water and}$ $\text{surroundings and time as shown in the figure. The initial temperature}$ $\text{of the water is taken as } 80^\circ \text{C. The value of } t_2 \text{ as mentioned in the}$ $\text{graph will be:}$

Options:

1. $12$
2. $14$
3. $16$
4. $18$

Solution:

$\text{Hint: Use Newton's law of cooling.}$ $\text{Step 1: Find the value constant.}$ $\text{Apply Newton's law of cooling,}$ $-\frac{dQ}{dt} = k(T_m - T_s)$ $\text{In the first case,}$ $dQ = -20, \ T_m = \frac{80 + 60}{2} = 70^\circ \text{C}, \ dt = 6 \ \text{min}, \ T_s = 20^\circ$ $\Rightarrow \frac{20}{6} = k(70 - 20) \Rightarrow k = \frac{1}{15}$ $\text{Step 2: Find the value of time } t_2.$ $\text{In the second case,}$ $T_f = 40^\circ, \ T_i = 60^\circ, \ T_m = \frac{60 + 40}{2} = 50^\circ, \ T_s = 20^\circ$ $\Rightarrow \frac{20}{t_2 - 6} = \frac{1}{15}(50 - 20) \Rightarrow t_2 - 6 = 10 \Rightarrow t_2 = 16 \ \text{min}$ $\text{Hence, option (3) is the correct answer.}$

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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
}