Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Answer:
Q = 114349.5 J
Explanation:
Hello there!
In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:
[tex]Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J[/tex]
Thus, the total energy turns out to be:
[tex]Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J[/tex]
Best regards!
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.