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Current Question (ID: 11099)

Question:

$\text{A Carnot engine whose sink is at 300 K has an efficiency of 40\%. By how much should the temperature of the source be increased to increase its efficiency by 50\% of its original efficiency?}$

Options:

1. $275 \text{ K}$
2. $325 \text{ K}$
3. $250 \text{ K}$ (Correct)
4. $380 \text{ K}$

Solution:

$\text{Hint: } \eta = 1 - \frac{T_2}{T_1}$ $\text{Step 1: Find the initial temperature of the source.}$ $\text{The efficiency of the Carnot engine is given by;}$ $\eta = 1 - \frac{T_2}{T_1}$ $\Rightarrow 0.40 = 1 - \frac{300}{T_1}$ $\Rightarrow \frac{300}{T_1} = 1 - \frac{40}{100}$ $\Rightarrow \frac{300}{T_1} = \frac{60}{100}$ $\Rightarrow T_1 = T_i = 500 \text{ K}$ $\text{Step 2: Find the final temperature of the source.}$ $\text{The efficiency of the Carnot engine is given by;}$ $0.6 = 1 - \frac{300}{T_1} \quad [\eta = 0.6 \text{ (0.4 + 0.2 = 0.6)}]$ $\Rightarrow \frac{300}{T_1} = 1 - \frac{60}{100}$ $\Rightarrow \frac{300}{T_1} = \frac{40}{100}$ $\Rightarrow T_1 = T_f = 750 \text{ K}$ $\text{Step 3: Find the increase in the temperature of the source.}$ $\text{The increase in the temperature of the source is given by;}$ $\Delta T = T_f - T_i$ $\Rightarrow \Delta T = 750 - 500 = 250 \text{ K}$ $\text{Hence, option (3) is the correct answer.}$

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
}