1.7 x 10 8 J of energy is removed from 60 kg of steam initially at 260°F. You ma

Question

1.7 x 108 J of energy is removed from 60 kg of steam initially at 260°F. You may assume that the steam is cooled at constant pressure and that the specific heat of the steam is 1850 J/kg°C. What is the final temperature? What is the final phase? 1.7 x 108 J of energy is removed from 60 kg of steam initially at 260°F. You may assume that the steam is cooled at constant pressure and that the specific heat of the steam is 1850 J/kg°C. What is the final temperature? What is the final phase? What is the final temperature? What is the final phase?

Explanation / Answer

Heat removed Q =1.7 x 108 J
Mass of steam m = 60 kg Initial temperature t = 260°F                                = [(5/9)(F-32)]                                = ( 5/ 9 ) (260-32)                                = 126.666 o C Amount of heat removed when it cool to 100 o C steam Q ' = mC (126.666-100) Where C = The specific heat of the steam                = 1850 J/kg°C. So, Q' = 2.96 x 10 6 J Amount of heat removed when it is converted into 100 o C steam to 100 o C water Q " = mL Where L = Latent heat of vaporisation                = 2.26 x 10 6 J So, Q " = 135.62 x 10 6 J Amount of heat removed when water at 100 o C is converted into water at 0 o C is              Q "'= mC '( 100- t ) Where C ' = Specific heat of water                   = 4186 J / kg o C So, Q"' = 25.116 x 10 6 J Q ' + Q" +Q"' = 1.6369 x 10 8 J Final temperature = 0 o C Final phase : water - ice phase          

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