Homework Help: Chemistry

Posted by Casey on Friday, July 2, 2010 at 12:57pm.

I am having a horrible time with Enthalpy change. I understand how to find specific heat I believe, but the heat capacity of the vessel is given as 10 J/K. My problem is that my professor did not show us how to work this type of problem, and I desperately need help with it. Here is the following problem. I am trying to find the final temperature of the system. I feel like I am missing something, PLEASE HELP!

A 670. g piece of copper tubing is heated to 97.1°C and placed in an insulated vessel containing 57.5 g of water at 40.0°C. Assuming no loss of water and heat capacity for the vessel of 10.0 J/K, what is the final temperature of the system (c of copper = 0.387 J/g·K)?

Chemistry - DrBob222, Friday, July 2, 2010 at 1:18pm
heat lost by Cu + heat gained by water + heat gained by calorimeter = 0

[massCu x specific heat Cu x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] + [10 x (Tfinal-Tinitial)] = 0
Solve for Tfinal, the only unknown.
Chemistry - Casey, Friday, July 2, 2010 at 7:20pm
Thankyou, that helped so much.

Answer this Question

Related Questions

Lauren - The amount of heat required to raise temperature of 55.85 g of iron 1 ...
Chemistry - when 5.0g of KOH is added to 100mL, of water, the temperature of ...
chemistry - At standard pressure, ammonia melts at 195 K and boils at 240 K. If ...
chemistry - If someone could see if I have the right answers, and if I don't...
intro chem HELp please - 1. Zinc is often used to calibrate calorimeters because...
chemistry - 3. Based on your answer in question 2, determine the percent error ...
chem - At standard pressure, ammonia melts at 195 K and boils at 240 K. If a ...
Chemistry - Which of the following is the best definition of Hess' Law? A. ...
Chemistry - the specific heat capacity of gold is .13 J/g°C. Calculate the ...
Chemistry Help! - the specific heat capacity of gold is .13 J/g°C. Calculate...

For Further Reading

First Name:
School Subject:
Answer: