Answer:
<u>225.6 kJ</u>, <em>assuming the water is already at 100 °C</em>
Explanation:
The correct answer to this question will depend on the initial temperature of the water to which heat is added to produce steam. Energy is required to raise the water temperature to 100°C. At that point, an energy of vaporization is needed to convert liquid water at 100 °C to water vapor at 100°C. The heat of vaporization for water is 2256.4 kJ/kg. The energy required to bring 100g of water from a lower temperature to 100°C is calculated at 4.186 J/g°C. We don't know the starting temperature, so this step cannot be calculated.
<em><u>Assuming</u></em> that we are already at 100 °C, we can calculate the heat required for vaporization:
(100.0g)(1000.0g/1 kg)(2256.4 kJ/kg) = 225.6 kJ for 100 grams water.
Answer:
Q = 143,921 J = 143.9 kJ.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:
Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:
Regards!
"Endothermic" is the way among the following choices given that you would <span>classify the following chemical reaction, in terms of energy. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has come to your help.</span>
