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Current Question (ID: 19623)

Question:

$\text{Two identical beakers } A \text{ and } B \text{ contain equal volumes of two different liquids at } 60^\circ C \text{ each and left to cool down.}$ $\text{Liquid in } A \text{ has density of } 8 \times 10^2 \text{ kg/m}^3 \text{ and specific heat of } 2000 \text{ Jkg}^{-1}\text{K}^{-1}$ $\text{while liquid in } B \text{ has density of } 10^3 \text{ kgm}^{-3} \text{ and specific heat of } 4000 \text{ Jkg}^{-1}\text{K}^{-1}.$ $\text{Which of the following best describes their temperature versus time graph schematically?}$ $\text{(assume the emissivity of both the beakers to be the same)}$

Options:

1. $\begin{array}{c} T \\ 60^\circ C \end{array} \quad \begin{array}{c} A \text{ and } B \\ \quad \end{array} \quad t$
2. $\begin{array}{c} T \\ 60^\circ C \end{array} \quad \begin{array}{c} B \\ A \end{array} \quad t$
3. $\begin{array}{c} T \\ 60^\circ C \end{array} \quad \begin{array}{c} A \\ B \end{array} \quad t$
4. $\begin{array}{c} T \\ 60^\circ C \end{array} \quad \begin{array}{c} A \\ B \end{array} \quad t$

Solution:

$\text{Hint: } -\frac{dT}{dt} \propto \frac{1}{pS}$ $-\frac{dT}{dt} = \frac{\varepsilon \sigma A}{ms} 4T_0^3 \left( T - T_0 \right) \propto \frac{1}{pS}$ $\text{For } A \quad ps = 800 \times 200 = 16 \times 10^5$ $\text{For } B \quad ps = 10^3 \times 4000 = 40 \times 10^5$ $(ps)_B > (ps)_A = \left( -\frac{dT}{dt} \right)_B < \left( -\frac{dT}{dt} \right)_A$

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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
}