May 22, 2019
Just like all physics problems, calculations for latent energy require using the GRASP method.
G – List the Givens
R – List what is Required
A – Analysis – Analyze the problem
S – Solution
P – Paraphrase
When a car brakes, an amount of thermal energy equal to 112,500 J is generated in the brake drums. If the mass of the brake drums is 28.0 kg and its specific heat capacity is 460.5 J kg-1oC-1, what is the change in its temperature?
G– TE (112,500 J), Mass (28 kg), H-C (460.5 K kg)
R – Change in temp
A– Q=MC^T
S – 112,500 = 28 X 460.5 X ^T
112, 500 / 12,894 = ^T
244.30 K = ^6
P – Therefore, the change in temperature in this case is 244,30 K
After Unit Reflection: In this post, I demonstrated an understanding of latent energy transformations and problems. I showed the GRASP solution, and what each letter stands for, and then showed an example calculations.
Energy E Blog 