Answer:
The final temperature at the equilibrium is 204.6 °C
Explanation:
Step 1: Data given
Mass of iron = 60.0 grams
Initial temperature = 250 °C
Mass of gold = 60.0 grams
Initial temperature of gold = 45.0 °C
The specific heat capacity of iron = 0.449 J/g•°C
The specific heat capacity of gold = 0.128 J/g•°C.
Step 2: Calculate the final temperature at the equilibrium
Heat lost = Heat gained
Qlost = -Qgained
Qiron = -Qgold
Q=m*c*ΔT
m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)
⇒with m(iron) = the mass of iron = 60.0 grams
⇒with c(iron) = the specific heat of iron = 0.449 J/g°C
⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C
⇒with m(gold) = the mass of gold= 60.0 grams
⇒with c(gold) = the specific heat of gold = 0.128 J/g°C
⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C
60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )
26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)
26.94T2 - 6735 = -7.68T2 + 345.6
34.62T2 = 7080.6
T2 = 204.5 °C
The final temperature at the equilibrium is 204.6 °C
