Question

A 6.40 g sample of iron (specific heat capacity =0.451 J/g*C) is placed in a boiling water bath until the temperature of the metal is 100.0*C. The metal is quickly transferred to 119.0g of water at 25.0*C in a calorimeter (specific heat capacity of water = 4.18 J/g*C). Determine the final temperature of the water in the calorimeter (3 significant figures).

Answer 1

This is an example for heat balance q = m x C_s x ΔT

Heat balance is: heat gained by 119.0g of water = heat lost by 6.40 g sample of iron

So m1 x C_w x ΔT = m2 x C_i x ΔT

Let t be the final temp of water mixture.

(119 g) ( t-25 ) (4.18 J/g*C) = (6.4g) (100-t) (0.451 J/g*C)

497.42 (t-25) = 2.8864 (100-t)

497.42t - 12435.5 = 288.64 - 2.8864t

500.3064t = 12724.14

Therefore t = 12724.14/500.3064 = 25.4326948446 ^oC = 25.4 ^oC

Answer : 25.4 ^oC (3 significant figures)

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