o/n 2005 for question 3b (i) why is work done negative, isnt work dont on the system?
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?