A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample starts off at 91.3°C and the temperature of the water starts off at 16.0°C. When the temperature of the water stops changing it's 19.5°C. The pressure remains constant at 1atm.

Calculate the specific heat capacity of aluminum according to this experiment.

Be sure your answer is rounded to the correct number of significant digits.

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

a small container that holds a source of energy a solar cell that captures light energy and converts it into electricity a voltaic cell that uses the oxidation of a fuel to produce electricity

explanation:

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.