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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Shweta Thakur posted an Question
September 24, 2020 • 01:36 am
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30 points
30 points
- IIT JAM
- Chemistry (CY)
What is the use of specific heat here btw i got 2 kj by direct subtracting -393-(-395) 98 eats of these substances are 720 and 505
Lal 20. The heats of combustion of graphite and diamond at 298 K are -393 and -395 kJ/mole respectively. The specific heats of these substances are 720 and 505 J kg Krespectively. Calculate the heat of transformation of graphite into 28. A heat pu 15°C and from the diamond at 273 K. (2-0645 kJ mole 29. Given the 12 [Hint: AH= [-393 -(395) }+ (0-720-0-505) 1000 x (298-273) ]
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Priyanshu kumar Best Answer
Here at different temperature it is asking for finding of heat of transformation. So it is used here,if it is asking only for same temperature given then you can proceed directly 👍
see the hint ..also and I am getting 2 answer right n
But ans is 2.0645 and you get only 2...
slightly difference occur here as only at diff temp it is asking
then plz expalin the answer
beacuse I have just subtracted the heat combustions
here Del H = m del Cp delT formula is used
For exact calculation you can use but for nearly you directly calculate
sir
ik Baar aap kr dena sir
when u get free
sir
okk wait
bs ise calculate krna tha
ok
this you got na or any doubt then please ask