Saturday

October 25, 2014

October 25, 2014

Posted by **A** on Wednesday, March 2, 2011 at 12:51am.

An object at 20 degrees Celsius with heat capacity 100 calories/Celsius is placed in 40 grams of water at 100 degrees Celsius. After equilibrium is reached, the object is removed from the hot water and placed in 200 grams of water at 0 degrees Celsius. What is the final temperature of the object? Explain.

Thanks for the help!

- Physics -
**drwls**, Wednesday, March 2, 2011 at 1:25amYou will have t do this problem in two steps.

In step 1, determine the equilibrium temperature when the object is placed in the 100 C water. The heat lost by the water will equal the heat gained by the object. Write that equation and solve for the final temperature of that step, T1.

(100 cal/C)*(T1-20)

= (40 Cal/C)*(100 - T1)

Solve for T1

(100-T1)/T1-20) = 2.5

100 - T1 = 2.5 T1 -50

3.5 T1 = 150

T1 = 42.8 C

Once you have T1, write an analogous equation for the process of reaching thermal equilibrium in the second step. Then solve for the final equilbrium temperature, T2.

**Answer this Question**

**Related Questions**

Physics - Please Help! An object at 20 degrees Celsius with heat capacity 100 ...

Physics - Please Help! An object at 20 degrees Celsius with heat capacity 100 ...

Physics - Please Help! An object at 20 degrees Celsius with heat capacity 100 ...

Physics - Please Help! An object at 20 degrees Celsius with heat capacity 100 ...

Physics - Please Help! I am having difficulties grasping these concepts. Explain...

Physics 20 - 1. How much heat energy is required to change 500 grams of water at...

chemistry - A 100.0 gram sample of iron at 80.0 degrees Celsius was placed in ...

Chemistry - 30 grams of a metal at 100 degrees Celsius is placed in a ...

Chemistry - 30 grams of a metal at 100 degrees Celsius is placed in a ...

science - a 100.0 gram metal sample is heated to a temperature of 110 degrees ...