Answer:
The specific heat of the metal is 0.34 J/g°C
Explanation:
Step 1: Data given
Mass of the metal = 12.0 grams
Mass of the water = 25.0 grams
Initial temperature of the metal = 90.0 °C
Initial temperature of the water = 22.5 °C
Final temperature = 25 °C
Specific heat of water = 4.184 J/g°C
Step 2: Calculate the specific heat of the metal
Qlost = Qgained
Q = m*c*ΔT
Qmetal = -Qwater
m(metal) *c(metal)* ΔT(metal) = -m(water) * c(water) *ΔT(water)
⇒ with mass of metal = 12.0 grams
⇒ with c(metal) = TO BE DETERMINED
⇒ with ΔT(metal) = T2 - T1 = 25.0°C - 90.0 °C = -65.0 °C
⇒ with mass of water = 25.0 grams
⇒ with c(water) = 4.184 J/g°C
⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C
12.0 * c(metal) * -65.0 °C = -25.0g * 4.184 J/g°C * 2.5°C
-780.0 * c(metal) = -2615 ( 2.6*10^3 with sig figs)
c(metal) = 0.335 (=0.34 with sig figs)
The specific heat of the metal is 0.34 J/g°C
