A 29.60 g piece of iron and a 28.00 g piece of gold at 100.0 C...

A 29.60 g piece of iron and a 28.00 g piece of gold at 100.0 C were dropped into 880.0mL of water at 17.60 C. The molar heat capacities of iron and gold are 25.19 J/(mol*C) and 25.42 J/(mol*C), respectively. What is the final temperature of the water and pieces of metal?

Expert Solution

Moles of gold = mass/ molar mass of gold = 28 /196.97 = 0.1423736

Moles of iron = 29.6 /55.845 = 0.53

Heat lost by gold = Molar heat of gold x moles of gold x temp change

= 25.42 x 0.1423736 x ( 100-T)

= 3.61914 ( 100-T)

Heat lost by Iron = Molar heat of iron x moles of iron x temp change

= 25.19 x 0.53 x ( 100-T) = 13.3507 ( 100-T)

Heat gained by water = specific heat of water x mass of water x temp change

= ( 4.184 J/gK) x ( 880 g) x ( T-17.6) ( water has dnesity 1g/ml hence 880 ml =880g)

= 3681.92 ( T-17.6)

Now heta lost by iron + heta lost by gold = heta gained by water

13.3507 ( 100-T) + 3.61914( 100-T) = 3681.92 ( T-17.6)

1335.07 - 13.3507T + 361.914 -3.61914T = 3681.92 T - 64807.79

66504.774 =3699 T

T =18 C is the final temp

queen_honey_blossom answered 2 years ago

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