Mass of aluminum 16.27g Initial temperature of Al 83.85°C
Volume of water 51.8ml
Initial temperature of water 20.30°C
Final water temperature 23.90
Density of water 1.00 g/ml
Specific heat of water 4.184 J/g.°C
Calorimeter constant 22.44 J/°C

A student adds a heated sample Of pure aluminum metal to a Styrofoam coffee cup calorimeter containing deionized water. Use the collected data to answer the following questions.

Required:
Assuming that heat was transferred from the aluminum to the water and the calorimeter, determine the specific heat of aluminum.

yes

Explanation:

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