Answer:
[tex]Q=-126.1kJ[/tex]
Explanation:
Hello,
In this case, by means of the released heat, we need to consider the cooling of water in two steps:
1. Condensation of steam at 100 °C.
2. Cooling of water from 100 °C to 37 °C.
Therefore, we need the enthalpy of condensation of water that is 40.65 2258.33 J/g and the specific heat that is 4.18 J/g°C for the same amount of cooled water to obtain:
[tex]Q=50.0g*[-2258.33\frac{J}{g}+4.18\frac{J}{g\°C}(37-100)\°C]\\\\Q=-126.1kJ[/tex]
Best regards.
