Question

If 1.10 g of steam at 100.0 °C condenses into 38.5 g of water, initially at 27.0 °C, in an insulated container, what is the final temperature of the entire water sample? Assume no heat is lost to the surroundings.

Answer 1

Temperature = 44.02°C

Step-by-step explanation:

Insulated container indicates no heat loss to the surroundings.

The specific heat capacity of a substance is a physical property of matter. It is defined as the amount of heat that is to be supplied to a unit mass of the material to produce a unit change in its temperature.

The SI unit of specific heat is joule per kelvin and kilogram, J/(K kg).

Now,

specific heat for water is 4.1813 Jg⁻¹K⁻¹.

Latent heat of vaporization of water is 2257 Jg⁻¹.

Energy lost by steam in it's process of conversion to water, is the energy acquired by water resulting in an increase in it's temperature.

`Q = mS \Delta T`

Q= Heat transferred

m= mass of the substance

T= temperature

Also,

`Q = mL`

L= Latent heat of fusion/ vaporization ( during phase change)

Now applying the above equations to the problem:

`m_(w) S_(w) (T-27) = (m_(s) L) + m_(s)S_(w) (100 -T)`

`38.5 * 4.1813 * (T-27) = (1.10 * 2257) + 1.10 * 4.1813 * (100 -T)`

Temperature = 44.02°C