Open Question
A Carnot engine whose hot-reservoir temperature is 400℃ has a thermal efficiency of 40%. By how many degrees should the temperature of the cold reservoir be decreased to raise the engine's efficiency to 60%?
23. The Second Law of Thermodynamics
The Carnot Cycle
Struggling with Physics?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
A Carnot engine with an efficiency of 70% is cooled by water at 10°C. What temperature must the hot reservoir be maintained at?
A
33.3 K
B
404.3 K
C
943.3 K
D
14.3 K
Verified step by step guidance1
Understand the concept of a Carnot engine: A Carnot engine is an idealized heat engine that operates between two thermal reservoirs, a hot reservoir and a cold reservoir, with maximum efficiency determined by the temperatures of these reservoirs.
Recall the formula for the efficiency of a Carnot engine: Efficiency (η) is given by η = 1 - (T_cold / T_hot), where T_cold and T_hot are the absolute temperatures of the cold and hot reservoirs, respectively.
Convert the temperature of the cold reservoir from Celsius to Kelvin: Since the cold reservoir is cooled by water at 10°C, convert this to Kelvin by adding 273.15, resulting in T_cold = 283.15 K.
Set up the equation using the given efficiency: With an efficiency of 70%, or 0.70, the equation becomes 0.70 = 1 - (283.15 / T_hot).
Solve for T_hot: Rearrange the equation to find T_hot by isolating it on one side, which involves solving T_hot = 283.15 / (1 - 0.70).
6:28mWatch next
Master The Carnot Cycle and Maximum Theoretical Efficiency with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
The Carnot Cycle practice set
Problem sets built by lead tutorsExpert video explanations