[achibaby1974]

A system releases 126 kJ of heat while 140. kJ of work is done on it. Calculate E.

ok this is just -126 kJ + 140 = 14 right?

but why doesn't this go to 3 sig fig? why wouldn't it be 14.0? i also tried 1.40e1. i haven't tried 14 b/c the problem specifically shows 3 sig figs. so is it just 14? if so, why? i don't understand. 

ok. here's another one w/ specific heat.
Calculate q when 17.6 g water is heated from 28°C to 90.0°C.
so i do, 17.6g X 4.184 J/gK X 62 = 4565.5808

i did 4.6e3 and 4.57e3 and with both answers it says: 
Your answer is off by a multiple of ten

what in the world does this mean?

[Yggdrasil]

The rules for sig figs with addition/subtraction are different for the rules for sig figs with multiplication/division.  The answer will have two sig figs.

For your second problem, do you have to input your answer with any specific unit (e.g. J or kJ)?

[achibaby1974]

For your second problem, do you have to input your answer with any specific unit (e.g. J or kJ)?
[/quote]

nope. it just says my answer is off by a multiple of ten. however it provides that i have to put my answer in kJ. wait.... that means that i have to divide by a thousand doesn't why? i'd have 2 sig figs right?

i'll give you the ex. we did in lecture yesterday calculate q for 100g of Hg that goes from 20 C to -39 C. the spec. ht of Hg is .140 according to our teacher. so, then we just multipy through                              100.0 X -59 x .140 = -827J. but here we didn't divide anything.

also when i googled water spec ht. I got 4.184 J/gC and in our book it's also 4.184 J/g C. why does the C and K not matter. where does the kJ and J come into play?