[Shea]

If 200g of water initially at 50°C has 1000J of heat removed from it, what is the final temperature of the water?

What's the formula for this kind of question?

[xiankai]

? Q = m c ? T

where

Q = amount of heat energy
m = mass of substance
c = specific heat capacity of substance
T = temperature

[skyglow1]

The triangle in xiankai's equation mean's "change in" if you did not know. It is read as "delta" so for the last number you are putting in the change in temperature or "delta T". You should be able to work it out now

[xstrae]

by the way, how do you type the 'triangle' or 'delta'. also, how do you type the symbol for wavelength? just thought i will ask since you are on it

[xiankai]

i go to microsoft word and insert the symbol i want, then paste it here. hehe

actually, you can use character map in windows too, or experiment with Alt - Numpad Keys (a list can be found in the shortcut keys for character map/MS word symbols)

for example,
Alt - 6 - 5 - 4 gives me Ä
Alt - 6 - 4 - 5 gives me à
Alt - 9 - 5 - 1 gives me ?

be sure to keep your numlock key on so you don't activate unnecessary commands like Home, PageUp, PageDown, End, Insert.

if u're more interested, google for character map

[Donaldson Tan]

For Win95 and above, character map can be founder under Start->Programs->Accessories

Else type "charmap" in Start->Run, then press enter. This will run Character Map

[kayamusty]

Q=m.c ? T
Cal. =g. cal/g⁰C .⁰C

you should convert joule to cal.  and C_water=1 cal/g ⁰C

[english]

Just remember that heat is conserved.  I'm assuming this problem is at constant pressure.

So your heat that was removed from the water (system) will be absorbed by the surroundings, which in this case is whatever is holding the water.

Your temperature differences should reflect this notion; so basically since the water has lost heat, its temperature should be less than 50°C.

Also make sure that you understand that in this case whatever you are given in Celsius is also the same for Kelvin.  So your 50°C = 50 K.  Just in the case of these types of problems the scale is like this.  Do the math and you'll realize why.

[Borek]

Also make sure that you understand that in this case whatever you are given in Celsius is also the same for Kelvin.

Differences are identical, not temperatures itself.

[english]

Also make sure that you understand that in this case whatever you are given in Celsius is also the same for Kelvin.

Differences are identical, not temperatures itself.

Right.