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jannaleigh
14.04.2020 •
Chemistry
A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.0 °C are brought into contact with one another. Assuming no heat is lost to the surroundings, what will be the temperature when the two metals reach thermal equilibrium? The specific heat capacity of iron = 0.449 J/g•°C and gold = 0.128 J/g•°C.
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Ответ:
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
Ответ: