Answer:
The final temperature, at the equilibrium is 24.14 °C
Explanation:
Step 1: Data given
specific heat capacity of alloy = 0.260 J/(g°C)
MAss of alloy = 47 grams
Mass of water = 110 grams
Specific heat of water = 4.184 J/g°C
Initial temperature of water = 20.0 °C
Initial temperature of alloy = 180.0 °C
Step 2: Calculate the final temperature at equilibrium
Heat lost = heat gained
Qlost = -Qgained
Q(alloy) =- Q(water)
Q=m*c*ΔT
Q = m(alloy)*c(alloy)*ΔT(alloy) = -m(water) * c(water)* ΔT(water)
⇒with m(alloy) = the mass of alloy = 47.0 grams
⇒with c(alloy) = the specific heat of alloy = 0.260 J/g°C
⇒with ΔT(alloy) = the change of temperature = T2- T1 = T2 - 180 °C
⇒with m(water) = the mass of water = 110 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature = T2 - 20.0°C
47.0*0.260 * (T2 - 180.0) = - 110 * 4.184 * (T2 - 20.0)
12.22(T2-180.0) = -460.24(T2- 20)
12.22T2 - 2199.6 = -460.24T2 + 9204.8
472.46T2 = 11404.4
T2 = 24.14 °C
The final temperature, at the equilibrium is 24.14 °C
