Heat and Energy?

Question

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

How much energy is released when 1 L of O2 reacts with excess sulphur at STP?

I'd like to see how these are done. I'm lost.

Q = m*Cp*deltaT. I'm not looking for heat, I'm looking for the final temperature.

Answer 1

deltaQ=1000J
m=200g
Cp~1.87J/(g*K) assume constant
deltaT=(323.15K-Tf)

deltaQ=m*Cp*deltaT
1000J=(200g)*(1.87J/(g*K)*(323.15K-Tf)
1000/(200*1.87)-323.15=-Tf
Tf=320K of 47 deg C

Answer 2

There are a few ways to do this:

1 - Table lookup: look up the internal energy of water at 50º C, it will be in kJ/kg. Multiply that value by your mass of water to find initial internal energy. Subtract 1 kJ from that value. Use linear interpolation to determine the new temperature. Write your assumption of constant pressure. This is the most accurate because the Heat Capacity of a substance varies with temperature.
2. Apply a linear formula and use interpolated specific heat. The basic equation ∆U=mCp(T2-T1) yields T1=T2-(∆U/mCp). If you know your approximate temperature range then you can pick a Cp in the middle of it, and thats more accurate. A less accurate, but sometimes useful value is Cp at the known endpoint.

Note that The energy change as the liquid freezes (it might) isnt linear, and is very significant.

The second problem is a balance of energy:
Final Enthalpy of Formation = Initial Enthalpy of Formation + Energy Released.

Its very simple algebra.

Answer 3

For the water question, Q = m*Cp*deltaT

where Q is heat, m is mass or moles Cp is specific heat (J/mol-C) or heat capacity (J/g-C) and delta T is the change in temperature. So you can get final temperature from delta T, which is final temperature minus initial temperature..

For the O2 question Q is a function of the enthalpy of reaction. The exact equation will depend on what the enthalpy is in terms of (moles O2, Liters O2, grams H2O, etc...)