Question

33. An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6x 104 cal of heat at higher temperature. Amount of heat converted
into work is
(a) 1.2 x 10'cal
(b) 2.4x10*cal
(a) 6.2 x 10*cal
(d) 4.8 x 104 cal
​

Answer 1

Given :

• T₁ = 227 °C = 227 + 23 = 250 K
• T₂ = 127 °C  = 127 + 23 = 150 K
• Q₁ = 6 x 10₄ Cal

To find :

• Work

Solution :

Carnot's Cycle

• Hypothetical cycle
• Most efficient engine
• All the processes are reversible.

As per the condition,

$\implies\boxed{\sf{\dfrac{Q}{T} = const }}$

$\implies\sf{\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1}}$

$\implies\sf{Q_2 = \dfrac{T_2}{T_1}\times Q_1}$

Now let's substitute the given values,

$\implies\sf{Q_2 = \dfrac{150}{250}\times 6 \times 10^4}$

$\implies\sf{Q_2 = 0.6 \times 6 \times 10^4}$

$\implies\sf{Q_2 = 3.6 \times 10^4 Cal }$

___________________________

Now,

$\implies\boxed{\sf{W = Q_1 - Q_2}}$

Now let's substitute the given values,

$\implies\sf{W = (6-3.6) \times 10^6}$

$\implies\sf{W = 2.4 \times 10^6 J}$

Hence,

The amount of heat converted into work is 2.4 x 10⁶ J.