Ask Question
9 December, 05:33

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.

+1
Answers (1)
  1. 9 December, 07:21
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers