Question

How much total energy is released to cool 28.3 g of steam (water vapor) at 100.0°C to liquid water at 25.5°C? Water has a heat of vaporization of ± 40.7 kJ/mol and a specific heat of 4.184 J/g°C.

Answer 1

`\Delta H=-72870J=-72.9kJ`

Step-by-step explanation:

Hello,

In this case, for the computation of the total energy, we must consider two processes:

1. Condensation of steam (heat of condensation is the negative of heat of vaporization).

2. Cooling of hot water (we use the specific heat of water).

Thus, we write:

`\Delta H=\Delta H_(cond)+\Delta H_(cooling)`

For each term, we have:

`\Delta H_(cond)=28.3g*(1mol)/(18g) *-40.7(kJ)/(mol) *(1000J)/(1kJ)= -63989.44J`

`\Delta H_(cooling)=28.3g*4.184(J)/(g\°C)*(25.5-100.0)\°C =-8880.54J`

Therefore, the total energy results:

`\Delta H=-63989.44J-8880.54J\\\\\Delta H=-72870J=-72.9kJ`

Regards.