Question

Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum (660°C) and the other a block of solid mercury (-38.9°C). The engine runs by freezing 1.40 g of aluminum and melting 15.4 g of mercury during each cycle. The heat of fusion of aluminum is 3.97 x 105 J/kg; the heat of fusion of mercury is 1.18 x 104 J/kg. What is the efficiency of this engine?

Answer 1

The efficiency of this engine is 55.4 %

Step-by-step explanation:

Step 1: Data given

Reservoir 1: molten aluminium = 660°C

Reservoir 2: solid mercury = -38.9 °C

Mass of aluminium = 1.40 grams

Mass of mercury = 15.4 grams

The heat of fusion of aluminum is 3.97 * 10^5 J/kg

The heat of fusion of mercury is 1.18 * 10^4 J/kg

Step 2: Heat required to melt 15.4 g of Mercury

Q = m * Lf

Q = 0.015 Kg * 1.18 * 10^4 J/kg

Q = 177 J

Step 3: Energy absorbed by 1.40 g of aluminum

Q = m*Lf

Q = 0.0014 Kg * 3.97 * 10^5 J/kg = 555.8 J

Step 4: The total work done by the engine:

Wengine = Qh – Qc

Wengine = 397 J- 177 J = 220 J

Step 5:

efficiency = Wengine / Qh

Efficiency = 220 J / 397 J *100% = 55.4 %

The efficiency of this engine is 55.4 %