Posted by natash on Monday, March 31, 2008 at 9:11pm.
heat lost by Cu + heat gained by water = 0
mass x specific heat Cu x delta T = q Cu.
mass x specific heat H2O x delta T = q H2O.
delta T is Tfinal-Tinitial. You are given the initial T for both Cu and water, Tf is the unknown for which you solve. You must look up the specific heats of Cu and H2O.
I have assumed in the above that the final T of the water is NOT greater than 100. If it is another approach must be used.
I'm a bit confused so is this the right way to set up the problem
heat lost by Cu+heat gained by water=0
5.26g(.385)(Tf-258C)+125g(4.18)(Tf-21C)=0 then you solve for Tf?
Yes, you have it set up correctly. Solve for Tfinal.
Related Questions
chemistry - If a 5.26-g sample of copper at 258 °C is placed in 125 mL of ...
chemistry - A 49g piece of ice at 0.0C is added to a sample of water at 9.0C. ...
Chemistry - A 110.-g sample of copper (specific heat capacity = 0.20 J/°C &...
physical science - a 1.0 kg sample of metal with a specific heat of 0.50KJ/kgCis...
Chemistry Help(Calorimetry) - I have a few questions to ask, since I was absent ...
Physics - A hot, just minted copper coin is placed in 112g of water to cool. The...
Pysics - A hot, just minted copper coin is placed in 112g of water to cool. The ...
Chemistry - A sample of iron was heated to 161.0°C, then placed into 338 g ...
chemistry - A 15.0 g of sample of silver is heated to 100.0 0C and then placed ...
chemistry - mass of crucible- 26.4574g mass of crucible and hydrated sample- 27....
For Further Reading
