Question

15. A Carnot engine takes 6000 cal of heat from a reservoir at 627°C and gives it to a sink at 27°C. The work
done by the engine is
(1) 2 kcal
(2) 3 kcal
☺ 4 kcal
(4) 8 kcal​

Answer 1

Explanation:

Given A Carnot engine takes 6000 cal of heat from a reservoir at 627°C and gives it to a sink at 27°C. The work  done by the engine is

• T1 = 627 + 273 = 900 K
• Q1 = 6 x 10^3 cal
• T 2 = 27 + 273 = 300 K
• From Carnot theorem we get
• Q1 / T1 = Q2 / T2
• Q 2 = T2/ T1 x Q1
• Q2 = 300 / 900 x 6 x 10^3
•      = 2 x 10^3 cal
• Work done = Q1 – Q2
•                  = 6 x 10^3 – 2 x 10^3
•                   = 3 x 10^3 cal
•                   = 3 K cal
•                  = 3  x 4.2 x 10^3
•                   = 12.6 x 10^3 J

Answer 2

Explanation:

answer answer is 12.6 into 10 to the power 6 Joule