A5.00-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.
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Ответ:
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
Ответ:
The answer to your question is: 37228.5 J
Explanation:
See the picture below