A 40 % efficient heat engine does mechanical work at the rate of 400 W. If it runs for an entir

step 1 In this problem we are given the percentage efficiency of heat engine is equal to 40 % or effici

A 40 % efficient heat engine does mechanical work at the...

A $40 \%$ efficient heat engine does mechanical work at the rate of $400 \mathrm{~W}$. If it runs for an entire day, find the heat removed from the hot reservoir and the heat transferred to the cold reservoir.

Key Concepts

Energy Conservation in Thermodynamic Cycles

The principle of energy conservation states that the energy absorbed by the engine must equal the sum of the work output and the energy expelled to the cold reservoir. In thermodynamic cycles, the heat taken from the hot reservoir is partially transformed into work, while the remainder is rejected to the cold reservoir. This balances the energy inputs and outputs in the cycle and is a direct application of the first law of thermodynamics.

Time Conversion in Energy Calculations

Energy calculations for engines often require converting time units to calculate the total energy produced or consumed. For example, running at 400 W for an entire day involves converting that day into seconds (86400 seconds) to determine the total energy involved. Accurate time conversion is necessary for applying power figures to get total energy or work.

Heat Engine Efficiency

The efficiency of a heat engine is defined as the ratio of the work output to the heat input from the hot reservoir. In this context, a 40% efficient engine converts 40% of the absorbed heat into useful mechanical work, meaning that the rest of the energy is not converted into work and must be expelled to the cold reservoir. This concept is fundamental in understanding the limitations of real-world engines and the second law of thermodynamics.

Mechanical Work and Power

Mechanical work is the energy transferred when an object is moved by a force, and power is the rate at which this work is done. In this problem, the engine does work at a rate of 400 W, meaning it produces 400 joules of work per second. This concept is critical for calculating the total work done over a period, as it relates instantaneous power to total energy output over time.

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