Heat Capacities

Moderators: Chem_Mod, Chem_Admin

Posts: 31
Joined: Fri Sep 25, 2015 3:00 am

Heat Capacities

Postby Rachael_1H » Wed Jan 13, 2016 1:11 pm

20 grams of Copper at 100 degrees celsius is put in a vessel containing 50.7 grams water at 22 degrees celsius. Assume no energy is lost to surroundings. When finding the final water temperature, why do you set the +(heat released)= -(heat absorbed)?

Why do you make the "heat absorbed" negative?

Ani Galfayan 1H
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

Re: Heat Capacities

Postby Ani Galfayan 1H » Wed Jan 13, 2016 2:39 pm

In a perfect system q(sys) + q(surr) = 0. This occurs when there is no loss of heat. The surrounding "gains" what the system "loses" and the surrounding "loses" what the system "gains". (heat lost) + (heat gained) = 0
When you rearrange that equation you get q(sys) = -q(surr) which means that
(heat lost by the metal)= - (heat gained by water).
To explain the negative in terms of thermodynamics; when heat is lost the value is negative, so to make the equation true, a negative is placed where the heat is gained.
(- value) = - (+ value)

Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests