Question

How much energy is released or absorbed when 40.0 g of steam at 250.0 c is converted to water at 30.0 c?

Answer 1

The steam needs to condense, which occurs at 100C- 1. Q = (40.0g)(2.031J/gC)(250C-100C) The phase change is occurs- 2. Q = (40.0g)(79.72 cal/g) The water cools from 100C to 30C 3. Q = (40.0g)(4.18J/gC)(100C-30C)

Answer 2

Answer : The amount of energy released are, -36836.8 J

Solution :

Formula used :

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

where,

Q = heat or energy released or absorbed = ?

m = mass of water = 40 g

c = specific heat of water =
`4.186J/g^oC`

`\Delta T=\text{Change in temperature}`

`T_(final)` = final temperature =
`30^oC`

`T_(initial)` = initial temperature =
`250^oC`

Now put all the given values in the above formula, we get

`Q=40g* 4.186J/g^oC* (30-250)^oC`

`Q=-36836.8J`

The negative energy means energy is released and positive energy means energy is absorbed.

Therefore, the amount of energy released are, -36836.8 J