How many calories of energy are given off to lower the temperature of 100.0g of iron from 150.0°c to 35.0°c?

ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C

given the specific heat of iron as 0.108 cal/g·°C

heat=(100.0 g)(0.108 cal /g· °C )(125°C) =

  100x 0.108x125= 1350 cal

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

2b2t

explanation:

2b2t