calcuate the of heat of librated when 500gms of water at 100degree clesius is converted to water at 0 degree clecius
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Answer:
To convert 100.0 g of water at 20.0 °C to steam at 100.0 °C requires 259.5 kJ of energy.
Explanation:
This is like the Socratic problem here.
For this problem, there are only two heats to consider:
q1 = heat required to warm the water from 20.0 °C to 100.0 °C.
q2 = heat required to vapourize the water to steam at 100 °C.
q1=mcΔT=100.0 g × 4.184 J∘C−1g−1×80.0∘C=33 472 J
q2=mΔHvap=100.0 g × 2260 J⋅g−1=226 000 J
q1+q2=( 33 472 + 226 000) J = 259 472 J = 259.5 kJ(4 significant figures)
To convert 100.0 g of water at 20.0 °C to steam at 100.0 °C requires 259.5 kJ of energy.
Explanation:
This is like the Socratic problem here.
For this problem, there are only two heats to consider:
q1 = heat required to warm the water from 20.0 °C to 100.0 °C.
q2 = heat required to vapourize the water to steam at 100 °C.
q1=mcΔT=100.0 g × 4.184 J∘C−1g−1×80.0∘C=33 472 J
q2=mΔHvap=100.0 g × 2260 J⋅g−1=226 000 J
q1+q2=( 33 472 + 226 000) J = 259 472 J = 259.5 kJ(4 significant figures)
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