Question

An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 10⁴ cals of heat at higher temperature. Amount of heat converted to work isa) 4.8 × 10⁴ cal(b) 6 × 10⁴ cal(c) 2.4 × 10⁴ cal(d) 1.2 × 10⁴ cal

Answer 1

Work Input (or Heat Absorbs) = 6 × 10⁴ Cal.

T₁ = 127° C

 =(127 + 273) K. [Changing into kelvin]

 = 400 K.

T₂ = 227° C

 = (227 + 273) K. [Changing into kelvin]

 = 500 K.

∵ Efficiency = 1 - T₁/T₂

    = 1 - 400/500

    = (500 - 400)/500

   = 100/500

   = 1/5

Also, Efficiency = Work Output/Work Input

∴ 1/5 = Work Done/6 × 10⁴

∴ Work Done by the Carnot Engine = 60000/5

⇒ Work Done =  12000 Cal.

∴ Amount of Heat Converted into Mechanical Energy is 12000 Cal.

Note ⇒ The answer may be little bit different then the exact answer because I have taken the T₁ as 127 °C. It may be possible that in the book the value of Temperature is different. But the Methods of the Solution will be the same.So, you can change the Value of Temperature and apply the same method If the value in the question is different.

Hope it helps.☺️☺️⭐✨✨⭐✌️✌️❤️❤️❤️

Answer 2

For the largest mutual inductance, the two loops should be placed coaxially. In this case, flux through a loop due to another loop is the largest; hence, mutual inductance is the largest.

(b) For the smallest mutual inductance, the two loops should be placed such that their axes are perpendicular to each other. In this case, flux through a loop due to another loop is the smallest (