Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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!
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.