00:00рдирдореНрд╣реЗрд░ рдХрд▓ рдирдордмрд░ 2 рдЕрдл рдпреБрдирд┐рдЯ 17 рд╕реЗрдХрдВрдб рд▓рд╛ рдлреНрд░рдореЛ рдбрд╛рдпрдирдорд┐рдХреНрд╕ рдЗрд╕ рд╡рди рдХрд┐рд▓реЛ рдЧреНрд░рд╛рдо рдЖрдл рдЖрдЗрд╕ рдПрдЯ рдЬреАрд░реЛ рдбрд┐рдЧреНрд░реА рд╕реЗрдВрдЯрд┐рдЧреНрд░реЗрдб рдЗрд╕ рдореЗрдВ рдХреЗрдбрд╝рд╛ рд╣реИрдВ рдХрд┐ рдЖрдЗрд╕ рдЬреАрд░реЛ рдбрд┐рдЧреНрд░реА рд╕реЗрдВрдЯрд┐рдЧреНрд░реЗрдб рдХрдВрдкреНрдпреВрдд рдЗрдЯ рдЪреЗрдВрдЬ рдЗрди рдПрди рдПрдВрдЯреНрд░рд╛рдкреА
00:26рдЗрд╕ question рдореЗрдВ рдХрд╣ рд░рд╣рд╛ рд╣реИ рдХрд┐ ice 0 degree centigrade рдкрд░ melt рд╣реЛ рд░рд╣реА рд╣реИ рдФрд░ water рдореЗрдВ convert рд╣реЛ рд░рд╣реА рд╣реИ 0 degree centigrade рдкрд░
00:36so рд╣рдореЗрдВ рдХреНрдпрд╛ find рдХрд░рдирд╛ рд╣реИ we have to calculate change in entropy
00:41so first of all we will make a data
00:45as we know that mass of ice
00:52is given is one kilogram
00:57temperature of ice is also given which is 0 degree centigrade as we know that 0 degree centigrade
01:07is also equal to 273 Kelvin рдФрд░ latent heat of fuel
01:15рд╡реЛ constant рд╣реИ рд╡реЛ рдХрд┐рддрдиреА рд╣реИ latent heat of fuel
01:24is given means constant which is 3.36 into 10 raised power 5 joule рдкрд░ kg
01:35or we can also write this as рдЬрдм рд╣рдо 10 raised power 5 3.36 рдХреЗ рд╕рд╛рде
01:44we can reply рдХрд░реНрд╡рд╛рдПрдВрдЧреЗ so we will get 33.600 Joule рдкрд░ kg
01:59means 3 leg 36.000 Joule рдкрд░ kg and we have to find change in entropy means delta S
02:12means delta S so for solution or calculation
02:20we will use formula is change in entropy delta S is equal to delta Q upon T
02:33рдпрд╣рд╛рдБ рдкрд░ рджреЗрд▓реНрдЯрд╛ рдХреНрдпреВ рд╣рдореЗрдВ рдЖрдкреНрдЯреЗрди рдирд╣реАрдВ рд╣реИ рдЗрд╕ latent heat of fuel
02:42is equal to
02:43heat energy divided by mass
02:49so delta Q will be equal to
02:57mass into latent heat of fuel
03:01рдХреНрдпреЛрдВрдХреЗ рдпрд╣рд╛рдБ рдкрд░ mass divide рдХрд░ рд░рд╣рд╛ рд╣реИ рддреЛ рдмрд░рд╛рдмрд░ рдХреА рджреВрд╕рд░реА side multiply рдХрд░реЗрдЧрд╛
03:06so delta Q рдХрд╛ value рд╣рдо рдпрд╣рд╛рдБ рдкреБрдЯ рдХрд░реЗрдВрдЧреЗ
03:09so change in entropy will be equal to
03:13mass into latent heat of fuel divided by
03:18temperature
03:20now by putting the given values
03:24is
03:27mass is given which is
03:301 kg
03:32multiplied with latent heat of fuel of ice
03:37which is
03:40336
03:43000
03:44divided by
03:49temperature
03:53temperature is given which is
03:55273 Kelvin
03:58so change in entropy will be obtained as
04:01is
04:02рдЬрдм рд╣рдо рдЗрдирдХреЛ simplify рдХрд░реЗрдВрдЧреЗ
04:07so рд╣рдореЗрдВ final answer рдорд┐рд▓реЗрдЧрд╛
04:091230.76
04:16Joule
04:18per Kelvin
04:20so this is our
04:24final answer
04:29path
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