rasen58 Posted March 12, 2015 Posted March 12, 2015 There are two equal masses of silver insulated from each other. Sample A is maintained at 20 degrees C. Sample B has twice as much thermal energy as Sample A. Silver has a specific heat of 233 J/g*C. What is the temperature of sample B? I don't have any idea how to do this and don't have that good of a grasp of the term 'thermal energy'. I think thermal energy is equivalent to total kinetic energy of all the particles, so maybe you could use the formulas for average kinetic energy, but I'm not sure how. q = cm dT KEavg = (1/2)mv2 = (3/2)KT
Sensei Posted March 12, 2015 Posted March 12, 2015 Silver has a specific heat of 233 J/g*C. Not C but K rather.
swansont Posted March 12, 2015 Posted March 12, 2015 The thermal energy is the energy you could extract if you could cool it to absolute zero. U = cT Not C but K rather. Doesn't matter, since it's used for changes in temperature.
studiot Posted March 12, 2015 Posted March 12, 2015 I am not sure what level you are working at or why you have been given the specific heat. Here is a simple definition of Thermal Energy which leads to a simple equation, where the specific heats cancel in your case and results in an answer of 313oC http://www.answers.com/Q/What_is_The_thermal_energy_of_a_solid This definition ignores any expansion due to temperature rise. That is considers it insignificant. To include this you need the coefficent of thermal expansion of silver. If this is enough, post your working and we can discuss the difference between Thermal Energy, Internal Energy, Heat content, Heat capacity and their relationships to temperature, and the statement. The temperatures of both samples of silver lie between what is known as the Debye temperature of silver (-43oC) and the melting point of silver (960oC) We can discuss this as well if you like.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now