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.

Question

Mckayla Villa

Chemistry

23 April, 15:51

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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Laura Charles · Answer 1

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