Ask Question
7 July, 19:23

A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 j/°c · g) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 j/°c · g) are heated to 100.0°c. the mixture of hot iron and aluminum is then dropped into 93.1 g of water at 20.3°c. calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

+1
Answers (1)
  1. 7 July, 19:41
    0
    when (M*C*ΔT) Al + (M*C*ΔT) Fe = - (M*C*ΔT) w

    when water is gaining heat and Al& Fe losing heat

    when M (Al) = 5 g

    C (Al) = 0.89

    ΔT = 100 - Tf

    and when M (Fe) = 10 g

    C (Fe) = 0.45

    ΔT = 100 - Tf

    and Mw = 93.1 g

    Cw = 4.181

    ΔT = Tf - 20.3

    by substitution:

    ∴ 5 * 0.89 * (Tf-100) + 10 * 0.45 * (Tf - 100) = 93.1 * 4.181 * (Tf-20.3)

    ∴ Tf = 18.4 °C
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 j/°c · g) and a 10.00-g sample of iron pellets (specific heat capacity = ...” 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