a body along is moving a circular path of radius R what is its distance and displacement
Answers
Answer:
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
Step-by-step 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
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