Answer:
The specific heat of aluminium is 0.875 J/g°C
Explanation:
Step 1: Data given
The mass of the aluminium sample = 73.6 grams
Initial temperature of the sample = 95.0 °C
Mass of water = 100.0 grams
Initial temperature of water = 20.0 °C
Final temperature of water and aluminium = 30.0 °C
The specific heat of water = 4.184 J/g°C
Step 2: Calculate the specific heat of aluminium
Q gained = Q lost
Qwater = -Qaluminium
Q = m*c*ΔT
m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)
⇒ mass of aluminium = 73.6 grams
⇒ c(aluminium) = TO BE DETERMINED
⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C
⇒ mass of water = 100.0 grams
⇒ c(water ) = The specific heat of water = 4.184 J/g°C
⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C
73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C
-4784 * c(aluminium) = -4184
c(aluminium) = 0.875 J /g°C
The specific heat of aluminium is 0.875 J/g°C
