Question

A 28.8 g piece of iron and a 27.6 g piece of gold at 100.0°C were dropped into 660.0 mL of water at 16.0°C. The molar heat capacities of iron and gold are 25.19 J/(mol·°C) and 25.41 J/(mol·°C), respectively.

Answer 1

Given that;

28.8 g piece of iron and a 27.6 g piece of gold

The molar heat capacities of iron = 25.19 J/(mol·°C)

The molar heat capacities of gold = 25.41 J/(mol·°C),

Amount of water = 660.0 ml

Density of water is 1.0 g/ ml, then mass of water is 660 g

Temperature of metal = 100.0°C

Temperature of water = 16.0°C

We know that;

The molar heat capacities of water = 75.21J/mole-ºC

heat lost by metal = heat gained by water

first calculate the number of moles of both metal as follows:

number of mole = amount in g / molar mass

moles Fe = 28.8g / 55.85g/mole = 0.52 moles
moles Au = 27.6 g / 197g/mole = 0.14 moles
moles H2O = 660.0 g/ 18.02 g/ mole = 36.63
heat Fe + heat Au = heat water
(0.52 moles Fe x 25.19J/mole-ºC + 0.14 moles Au x 24.41J/mole-ºC)(100ºC - Tf) = (36.63 mole x 75.21J/mole-ºC)(Tf – 16.0 ºC)

13.1+ 3.42 (100 - Tf) = 2754.576 (Tf – 16.0)

16.52 (100 - Tf) = 2754.576 (Tf – 16.0)

1652 – 16.52 Tf = 2754.576 Tf – 44073.216

45725.216 = 2771.096 Tf

Tf = 45725.216 / 2771.096
Tf = 16.5 C