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!

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.