Answer:
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
