# [Thermal Physics] Calculating Energy Removal

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1. Feb 14, 2015

### Maxee

1. The problem statement, all variables and given/known data

A water of mass 300g has been poured into a glass of mass 100g (of specific heat capacity 970Jkg-1°C-1) in a refrigerator. How much energy must be removed to cool the liquid and the glass from 30°C to 7°C?

2. Relevant equations

c = Q/mΔT

3. The attempt at a solution

mwater = 300g = 0.3kg
cwater = 4180JKg-1°C-1
mglass = 100g = 0.1kg
cglass = 970Jkg-1°C-1
ΔT = -23°C
Q = ?

c = Q/mΔT => ΔT = Q/cm

...

I'm having a problem setting up the equation the right way since this isn't helping me solve my problem here. I believe I'm fundamentally thinking wrong about the problem and hence am setting it up this way. Any help would be appreciated on how to proceed with it and also, please, include the reasoning behind your decisions so I can learn to set up the problem correctly in my head. Thank you.

2. Feb 14, 2015

### Suraj M

split the question up, (treat the cup and the liquid independently), find the heat removed for the temperature change in each case and just add.
Try it out. Do the calculation and post it. You'll probably get it right!

3. Feb 14, 2015

### SteamKing

Staff Emeritus
It would be better to use Q = mcΔT in your calculations.

Qtotal = Qglass + Qwater

4. Feb 15, 2015

### Maxee

Thank you. I've done as I've been told and got both values for Q negative (since I included multiplication with -23 for ΔT in both equations). Now, when I turn Qwater to positive and than add Qglass to it [Q = Q + (-Q)] I get a good result. Would you please explain me what am I doing wrong, because I can see myself on the exam not realizing when I have to switch signs since there is no solution provided to compare my results.

5. Feb 15, 2015

### Staff: Mentor

The water and its container are both cooled, so for them ΔT is negative, and the energy (heat) change for them is likewise negative, i.e., Q is negative because their energy content must be reduced.

6. Feb 16, 2015

### Suraj M

Actually maxee, i prefer not using the signs at all, i would take even ΔT as positive, it makes calculations easy and takes the pressure off you, to get the right signs.
So just calculate the Q (change in energy) in both the cases and add ( only in this case, don't generalize), later see if its decrease or increase in temperature, then decide the sign.
I don't understand why you turned Qwater positive?? Even if you did then even Qglass should be turned +ve.