Question

50 grams of Lead at -10 C is placed in a container that has 200 grams of steam at 105 Celsius. What is the final temperature of the system? If not all the steam has condensed then state how much of it has condensed and how much remains in the form of steam.

Answer 1

Condensed steam, M = 0.091 g

The remaining steam = 199.9 g

Given:

mass of lead, m = 50 g

temperature, T = -10
`{^\circ}`

mass of steam, m' = 200 g

temperature, T' = 105
`{^\circ}`

Solution:

Now,

latent heat of vaporization of steam, L = 2260 J/g

specific heat of lead,
`C_(l)` = 0.128 J/g-K

specific heat of steam,
`C_(s)` = 1.996 J/g-K

Now, absorbed heat of lead to reach
`0^{\circ]`:

q = m
`C_(l)\Delta T`

q =
`50* 0.128* (100-(-10))`

q =
`50* 0.128* 110`

q = 704 J

Now, to reach 100
`{^\circ}` the amount of heat released by steam:

q' = m'
`C_(s)\Delta T`

q' =
`50* 1.996* (105 - 100)`

q' = 499 J

Now,

Let the condensed part of the steam have mass M, then:

ML = q - q'

M =
`(q - q')/(L)`

M =
`(704 - 499)/(2260)`

M = 0.091 g

The remaining steam = 200 - 0.091 = 199.9 g