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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.

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superman1987
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

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