Qualification > Sciences
Physics Paper4
Sue T:
--- Quote from: ruby92 on May 09, 2011, 07:55:55 pm ---
M/j 2005 5(b)
--- End quote ---
ur plottin the gradient of th given graph against time it should look like this
Sue T:
--- Quote from: ruby92 on May 09, 2011, 07:55:55 pm ---o/n 2005 for question 3b (i) why is work done negative, isnt work dont on the system?
--- End quote ---
ok so you calculated and its all fine - its not work done on the system - if you think about it the water or now steam would have 2 push against the atmospheric pressure - so its work done by the system - now:
delta U = delta W + delta Q
where U is internal energy, W is work done on the system, and Q is heat given to the system
notice how there aren't any negative signs in the equation
but when you come to calculate the last part, you cantttt take delta W as positive - since its by the system - therefore you put it in as : ''negative work done on the system'' = work done by system
get it?
Sue T:
--- Quote from: ruby92 on May 09, 2011, 07:55:55 pm ---M/j 2006 4(c), Q7 (a) and (b)
--- End quote ---
use ur imagination (though its physix)
elemis:
Good job Sue T. Keep it up and thanks for helping out our members !
ruby92:
Thank you :)
could you attached the working for the m/j 2004 question?
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