# Grade 11 Heat Capacity question

1. Dec 11, 2011

### Confuzed_one

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

a 0.20kg lead shot is heated to 90 degrees celsius and dropped into an ideal calorimeter containing 0.50kg of water initially at 20 degrees celsius. What is the final equilibrium temperature of the lead shot? The specific heat capacity of lead is 128 Joules/(kg degrees celsius) and the specific heat capacity of water is 4186Joules/(kg degrees Celsius)

Ml = 0.20
Cl = 128
Til = 90

Mw = 0.50
Cw = 4186
Tiw = 20

2. Relevant equations

Q= mass x heat capacity x difference in temperature

Qabsorbed + Qreleased = 0

3. The attempt at a solution

0.20 x 128 x (T2 - 90) + 0.50 x 4186 x(T2 - 20)

25.6 x (t2 -90) + 2093 x (T2-20)

I am completely stuck after this. I don't know what to do with the difference in temperature part. I've looked at example equations and it shows the temperature given getting transformed into joules? So in this equation the 90 and 20 would be replaced by different numbers with the unit Joules after. I don't understand why this is happening. Thank you for any help and I appreciate you taking the time to read this.

This may not be very clear, so if I need to explain more please let me know

Last edited: Dec 11, 2011
2. Dec 11, 2011

### Staff: Mentor

Make an equation out of it as in your Relevant Equations section. Solve for T2.

When units (Joules, °C, etc.) get confusing, take your equation and substitute each item with its corresponding units. Check how the units cancel and combine. They should all 'harmonize' to produce the expected units that are on the other side of the equation.

3. Dec 11, 2011

### Confuzed_one

I tried rearranging it to solve for T2, but I'm definitely doing something wrong.

0.20 x 128 x (t2 - 90)

-T2 = 0.20 x 128 - 90

-t2 = 25.6 - 90

-t2 = -64.4

T2 = 64.4

This is one of those multiple choice answer things, and this isnt an option sadly. I understand the whole unit thing, I'm just pressed for time and trying to shorten it. I really appreciate the help thank you

EDIT: I realized i should factor this, so I did it and I got the correct answer of 20.8 degrees celsius. I know this is kind of random and im not explaining myself, but you really helped and I got the right answer. Thank you very much, you have no idea how much i appreciate it

Last edited: Dec 11, 2011
4. Dec 12, 2011

### Delphi51

May I offer a little clarification at the beginning?
Heat lost by lead = heat gained by water
mC*ΔT = MC*ΔT
0.2*128*(90-T2) = .5*4186*(T2-20)