How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(c=0.895 jig cy from 22.5°c to 55 0°c? η ο ο ο οο 10x10) not enough information given prov 40 25 11 next >

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum =

= change in temperature

= initial temperature =

= final temperature =

Putting values in above equation, we get:

Therefore, the heat energy needed would be, 6486.5125 J

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j