Question

How much heat is released upon converting one mole of steam (18.0 g) from 100.0 ∘C to water at 25.0 ∘C? please show work and constants, trying to figure out how it works. only given the heat capacity for steam and water so try to only use that

Answer 1

Upon converting one mole of steam at 100.0 °C to water at 25.0 °C, the total heat released is the sum of the heat from condensation (40.7 kJ) and cooling (5.655 kJ), resulting in 46.355 kJ of heat released.

Step-by-step explanation:

To calculate the heat released when converting one mole of steam at 100.0 °C to water at 25.0 °C, we need to consider two steps: condensation and cooling. First, we calculate the heat released during the condensation of steam to water at 100.0 °C using the molar heat of vaporization, then we calculate the heat released during the cooling of water from 100.0 °C to 25.0 °C using the specific heat capacity for water.

The molar heat of vaporization of water is 40.7 kJ/mol. Thus, when 1 mole of steam condenses, the heat released is:
Heat released during condensation = 1 mole × 40.7 kJ/mol = 40.7 kJ

Next, to find the heat released during cooling, we use the specific heat capacity for water, which is approximately 4.18 J/g°C. The mass of one mole of water is 18.0 g, and the temperature change (ΔT) is 100.0 °C - 25.0 °C = 75.0 °C.

Heat released during cooling = mass × specific heat capacity × ΔT
= 18.0 g × 4.18 J/g°C × 75.0 °C
= 5655 J or 5.655 kJ

Finally, the total heat released is the sum of the heat from both processes:
Total heat released = Heat released during condensation + Heat released during cooling
= 40.7 kJ + 5.655 kJ
= 46.355 kJ

Therefore, 46.355 kJ of heat is released upon converting one mole of steam at 100.0 °C to water at 25.0 °C.

Answer 2

Answer: 46418.4 Joules

Step-by-step explanation:

The conversions involved in this process are :

`H_2O(g)(100^0C)\rightarrow H_2O(l)(100^0C)`

`H_2O(l)(100^0C)\rightarrow H_2O(l)(25^0C)`

Now we have to calculate the enthalpy change.

`Q=[[m* \Delta H_(condensation)]+[m* c_(p,l)* (T_(final)-T_(initial))]`

where,

Q = amount of heat released = ?

m = mass of water = 18 g

`c_(p,l)` = specific heat of liquid water =
`4.18J/g^oC`

`\Delta H` = enthalpy change = ?

m = mass of water = 25 g

`\Delta H_(condensation)` = enthalpy change for condensation = 2265 J/g

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

`Q=[(18g* 2265J/g)+(18g* 4.18J/g^oC* (100-25))^oC]`

`Q=40770+5648.4=46418.4J`

Therefore, the amount of heat released is 46418.4 J.