Answer:
The specific heat of the metal is 0.314 J/g°C
Explanation:
Step 1: data given
Temperature of the piece of metal = 100.0 °C
Mass of the metal = 85.5 grams
Volume of water = 122 mL = 122 grams
Temperature of water = 16.0 °C
The final temperature of water = 20.2 °C
The specific heat of water = 4.184 J/g°C
Step 2: Calculate the specific heat of metal
Heat gained= heat lost
Qgained = - Qlost
Qwater = -Qmetal
Q = m*c* ΔT
m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)
⇒m(metal) = mass of metal = 85.5 grams
⇒c(metal) = the specific heat of metal = TO BE DETERMINED
⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C
⇒m(water) = the mass of water = 122 grams
⇒c(water) = the specific heat of water = 4.184 J/g°C
⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C
85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2
c(metal) * (-6822.9) = -2143.9
c(metal) = 0.314 J/g°C
The specific heat of the metal is 0.314 J/g°C
