5. An aluminum cup of mass 46 g contains 150 g of water. The cup and water
are initially in thermodynamic equilibrium at room temperature (23 C). A 50-
g sample of an unknown alloy is heated to a temperature of 350 °C and
dropped into the water. The entire system (cup, water, alloy) is sealed in an
insulated container and allowed to come to equilibrium, at which point its
temperature is found to be 45°C. What is the specific heat of the alloy?
Hint: J
alloy - J + J Al cap

C(alloy) = 0.965 J/g°C

Explanation:

In the problem, the Heat of the alloy = - (Heat of water + Heat of Calorimeter). This, could be written as:

m×C(alloy)×ΔT = - ( m×C(H₂O)×ΔT +m×C(Al)×ΔT)

Replacing knowing C(H₂O) = 4.184 J/g°C and C(Al) = 0.900 J/g°C:

50g×C(alloy)×(45°C - 350°C) = - ( 150g×4.184J/g°C×(45°C - 23°C) +46g×0.900J/g°C×(45°C-23°C))

15250g°C×C(alloy) = 14718J

C(alloy) = 0.965 J/g°C

I hope it helps!

i dont know

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