Figure of speech used in the poem wall
Answers
Answer:
Follow me ☺️
Have a nice day ahead ❤️✔️
Explanation:
Converting 100. g of ice at 0.00 °C to water vapour at 100.00 °C requires 301 kJ of energy.
There are three heats to consider:
q1 = heat required to melt the ice to water at 0.00 °C.
q2 = heat required to warm the water from 0.00 °C to 100.00 °C.
q3 = heat required to vapourize the water to vapour at 100 °C.
q1=mΔHfus=100.g×334J1g = 33 400 J
q2=mcΔT=100.g×4.184J1°C∙g×100.00°C= 41 800 J
q3=mΔHvap=100.g×2260J1g = 226 000 J
q1+q2+q3 = 33 400 J + 41 800 J + 226 000 J = 301 000 J
= 301 kJ
the amount of heat energy absorbed by hundred grams of ice at zero degree centigrade to hundred degree centigrade steam : An object can have zero displacement even when it has moved a distance and shown some displacement. This takes place when the final position of an object coincides with its own initial position. Let's say, if a person moves around circular area and comes back to the place from where he started then the displacement will be zero.Square of velocity and position graph of a particle is given, at point A slope is 45" then acceleration of
particle at A ishttp://www.gowebrachnasagar.com/qrcode/cbse_cl10_science_exam.pdfA healthy dose of humor can also make a nice, uplifting message forsomeone you know well. “You mean so much to me. Hope you're feeling better very soon.” “Hope it helps a little to know how lovingly you're thought of.”
Converting 100. g of ice at 0.00 °C to water vapour at 100.00 °C requires 301 kJ of energy.
There are three heats to consider:
q1 = heat required to melt the ice to water at 0.00 °C.
q2 = heat required to warm the water from 0.00 °C to 100.00 °C.
q3 = heat required to vapourize the water to vapour at 100 °C.
q1=mΔHfus=100.g×334J1g = 33 400 J
q2=mcΔT=100.g×4.184J1°C∙g×100.00°C= 41 800 J
q3=mΔHvap=100.g×2260J1g = 226 000 J
q1+q2+q3 = 33 400 J + 41 800 J + 226 000 J = 301 000 J
= 301 kJConverting 100. g of ice at 0.00 °C to water vapour at 100.00 °C requires 301 kJ of energy.
There are three heats to consider:
q1 = heat required to melt the ice to water at 0.00 °C.
q2 = heat required to warm the water from 0.00 °C to 100.00 °C.
q3 = heat required to vapourize the water to vapour at 100 °C.
q1=mΔHfus=100.g×334J1g = 33 400 J
q2=mcΔT=100.g×4.184J1°C∙g×100.00°C= 41 800 J
q3=mΔHvap=100.g×2260J1g = 226 000 J
q1+q2+q3 = 33 400 J + 41 800 J + 226 000 J = 301 000 J
= 301 kJ