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

Answer:

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

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:

$\Delta H_{cond}=28.3g*\frac{1mol}{18g} *-40.7\frac{kJ}{mol} *\frac{1000J}{1kJ}= -63989.44J$

$\Delta H_{cooling}=28.3g*4.184\frac{J}{g\°C}*(25.5-100.0)\°C =-8880.54J$

Therefore, the total energy results:

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

Regards.