Answer:
The specific heat of the metal is 0.244 J/g°C
Explanation:
Step 1: Data given
Mass of metal = 150.0 grams
Temperature of the metal = 75.0 °C
Mass of water = 150.0 grams
Temperature of the water = 15.0 °C
Final temperature of the water = 18.3 °C
The specific heat of water is 4.184 J/g°C
Step 2: Calculate the specific heat of metal
Qlost = -Qgained
Q= m*C*ΔT
Qmetal = -Qwater
m(metal)*c(metal)*ΔT(metal) = - m(water)*c(water)*ΔT(water
⇒m(metal) = the mass of metal = 150.0 grams
⇒c(metal) = the specific heat of the metal = TO BE DETERMINED
⇒ΔT = The change of temperature = 18.3 - 75.0 = -56.7 °C
⇒m(water) = the mass of water = 150.0 grams
⇒c(water) = the specific heat of water = 4.186 J/g°C
⇒ΔT(water) = T2 - T1 = 18.3 - 15.0 = 3.3 °C
150.0 * c(metal) * -56.7 = -150.0 * 4.186 * 3.3
c(metal) = 0.244 J/g°C
The specific heat of the metal is 0.244 J/g°C
