Answer:
The specific heat of the metal is 0.335 J/g°C
Explanation:
<u>Step 1:</u> Data given
Mass of the metal = 12.0 grams
Initial temperature of the metal = 90.0 °C
Mass of the water = 25.0 grams
Initial temperature of water = 22.5 °C
Final temperature of water (and metal) = 25.0 °C
The specific heat of water = 4.18 J/g°C
<u>Step 2:</u> Calculate the specific heat of the metal
Qgained = -Qlost
Qwater = -Qmetal
Q= m*c*ΔT
m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)
⇒ mass of the metal = 12.0 grams
⇒ c(metal) = TO BE DETERMINED
⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0
⇒ mass of the water = 25.0 grams
⇒ c(water) = the specific heat of water = 4.18 J/g°C
⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C
12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5
c(metal) = 0.335 J/g°C
The specific heat of the metal is 0.335 J/g°C
