Answer:
The work done by this engine is 800 cal
Explanation:
Given:
100 g of water
120°C final temperature
22°C initial temperature
30°C is the temperature of condensed steam
Cw = specific heat of water = 1 cal/g °C
Cg = specific heat of steam = 0.48 cal/g °C
Lw = latent heat of vaporization = 540 cal/g
Question: How much work can be done using this engine, W = ?
First, you need to calculate the heat that it is necessary to change water to steam:

Here, mw is the mass of water

Now, you need to calculate the heat released by the steam:

The work done by this engine is the difference between both heats:
