One kilogram of ice at 0 degree centigrade is melted and converted to water at 0 degree celsius.Compute its change in entropy.

Physics with Akhtiar Ali  Bullo

One kilogram of ice at 0 degree centigrade is melted and converted to water at 0 degree celsius.Compute its change in entropy. Physics 12 class numerical solution  Inter class physics numerical solution  12 class physics numerical solution  Second law of thermodynamics numericals  New sindh text physics numericals

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00:00नम्हेर कल नमबर 2 अफ युनिट 17 सेकंड ला फ्रमो डायनमिक्स इस वन किलो ग्राम आफ आइस एट जीरो डिग्री सेंटिग्रेड इस में केड़ा हैं कि आइस जीरो डिग्री सेंटिग्रेड कंप्यूत इट चेंज इन एन एंट्रापी

00:26इस question में कह रहा है कि ice 0 degree centigrade पर melt हो रही है और water में convert हो रही है 0 degree centigrade पर

00:36so हमें क्या find करना है 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 और latent heat of fuel

01:15वो constant है वो कितनी है latent heat of fuel

01:24is given means constant which is 3.36 into 10 raised power 5 joule पर kg

01:35or we can also write this as जब हम 10 raised power 5 3.36 के साथ

01:44we can reply कर्वाएंगे so we will get 33.600 Joule पर kg

01:59means 3 leg 36.000 Joule पर 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

02:33यहाँ पर देल्टा क्यू हमें आप्टेन नहीं है इस 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

03:01क्योंके यहाँ पर mass divide कर रहा है तो बराबर की दूसरी side multiply करेगा

03:06so delta Q का value हम यहाँ पुट करेंगे

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

04:02जब हम इनको simplify करेंगे

04:07so हमें final answer मिलेगा

04:091230.76

04:16Joule

04:18per Kelvin

04:20so this is our

04:24final answer

04:29path

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