Question

6. Electrode potential of Zn?* /Zn is -0.76V and that of Cu?*/Cu is +0.34V. The
EMF of the cell constructed between these two electrodes is:
(a)+1.1V
(b)+0.42V
(c)-1.1V
(d)-0.42V​

Answer 1

Answer:

Given :

Heat gained = 4.5 × 10⁵ J

Heat released = 2 × 10⁵ J

To Find :

Percentage efficiency of steam engine and work done per cycle..

Solution :

❒ Efficiency of a heat engine/steam engine is given by, η = 1 - (Q₂ / Q₁)

Where,

η denotes efficiency

Q₂ denotes heat released to the sink

Q₁ denotes heat extracted from the source

By substituting the given values;

➛ η = 1 - (Q₂ / Q₁)

➛ η = 1 - (2 × 10⁵ / 4.5 × 10⁵)

➛ η = 1 - 0.44

➛ η = 0.56

Percentage efficiency = 56%

❒ Work done per cycle :

We know that work done is measured as the change in energy.

➠ W = Q₁ - Q₂

➠ W = (4.5 × 10⁵) - (2 × 10⁵)

➠ W = 2.5 × 10⁵ J