760 J = 32,174.76 kJThe power rating of heater is 15 kW, that means it will provide 15 kJ of energy every secondAssuming 100% efficiency, we have that no heat is lostTherefore, time required to provide the total energy = (32,......
= 22,570,000 J
Energy required to raise the temperature of steam from 373 K to 400 K = 10 * 1,996 * (400 - 373)
= 538,920 J
Therefore total energy required = 1,538,840 + 3,340,000 + 4,187,000 + 22,570,000 + 538,920
= 32,174,760 J
= 32,174.76 kJ
The power rating of heater is 15 kW, that means it will provide 15 kJ of energy every second
Assuming 100% efficiency, we have that no heat is lost
Therefore, time required to provide the total energy = (32,174.76 / 15)
= 2,145 s
= (2,145 / 60) min
= 35.75 minutes (Answer)
I hope you understand the above working. And I'm sorry for spelling and calculation mistakes, if any.
Here is the method.
Step 1. Change all temperatures to ºC (you could work in K, but ºC is probably slightly easier)
200K = 200-273 = -73ºC
Freezing/melting point of water = 0ºC
Boiling point of water = 100ºC
400K = 400-273 = 127ºC
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Step 2. Look up the following data if you were not given values. Make sure the values use J and kg.
c1 = specific heat capacity of ice
Lf - specific latent heat of fusion of ice
c2 = specific heat capacity of water
Lv = specific latent heat of vapourisation of water
c3 = specific heat capacity of steam (presumably at atmospheric pressure)
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Step 3. Work out the energy required.
Energy needed to heat ice to melting point = mc1(0 - (-73)) = 73mc1
Energy needed to melt ice = mLf
Energy needed to heat water from melting point to boiling point = mc2(100-0) = 100mc2
Energy needed to turn water to steam = mLv
Energy needed to heat steam to final temperature = mc3(127-100) = 27mc3
To simplify the arithmetic, note 'm' occurs in each expression, so the total energy,E is:,
E = m(73c1 +Lf + 100c2 + Lv +27c3) (in joules)
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Step 4. Find the time.
Since power = energy /time, time = energy /power
Time (in seconds) = E/15000
Convert this to minutes by dividing by 60.
Heat requirement, W for converting 1o kg of ice at 200 k to steam at 400 K is given by
W = 10*[2108*(273-200) + 80000*4.186+ 4186*(373-273)+540000*4.186+1996*(400-37… J or
W = 32216960 J = 3.22*10^7 J
time required by 15kW heater in minutes = [(3.22*10^7)/(15000*60)] = 35.79966 or 35.8 minutes
