Answer:
The specific heat of the metal is 0.289 J/g°C
Explanation:
Step 1: Data given
Mass of metal = 110.0 grams
Temperature of the metal = 82.00 °C
MAss of water = 110.0 grams
Temperature of the water = 27.00 °C
The final temperature = 30.56 °C
The specific heat of H2O is 4.18 J/(g°C).
Step 2: Calculate the specific heat of the metal
Heat lost = heat gained
Qlost = - Qgained
Qmetal = - Qwater
Q =m*c*ΔT
m(metal)*c(metal)*ΔT(metal) = -m(water) * c(water) *ΔT(water)
⇒with m(metal) = the mass of the metal = 110.0 grams
⇒with c(metal) = the specific heat of the metal = TO BE DETERMINED
⇒with ΔT(metal) = the change of temperature of the metal = T2 - T1 = 30.56 - 82.00 °C= -51.44 °C
⇒with m(water) = the mass of water = 110.0 grams
⇒with c(water) = the specific heat of water = 4.18 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 30.56 °C - 27.00 °C = 3.56 °C
110.0 * c(metal) * -51.44 = -110.0 * 4.18 * 3.56
c(metal) = 0.289 J/g°C
The specific heat of the metal is 0.289 J/g°C
