Question

A 94.7-g sample of silver (s = 0.237 J/(g · °C)), initially at 348.25°C, is added to an insulated vessel containing 143.6 g of water (s = 4.18 J/(g · °C)), initially at 13.97°C. At equilibrium, the final temperature of the metal–water mixture is 22.63°C. How much heat was absorbed by the water? The heat capacity of the vessel is 0.244 kJ/°C.

Answer 1

The piece of silver was initially at 348.25ºC. The piece of silver was dropped into an insulated vessel and the equilibrium temperature of the metal-water mixture in the vessel was 22.63ºC.

Heat lost by silver = (mass of silver)*(s, silver)*(change in temperature of the mixture)

= (94.7 g)*(0.237 J/g.ºC)*(348.25 – 22.63)ºC

= 7308.182718 J (keep guard digits extra).

The vessel as well as the water in the vessel gained heat.

Heat gained by the vessel = (Heat capacity of the vessel)*(change in temperature of the vessel)

= (0.244 kJ/ºC)*(22.63 – 13.97)ºC

= 2.11304 kJ

= (2.11304 kJ)*(1000 J)/(1 kJ)

= 2113.04 J.

As per the principle of thermochemistry,

Heat lost by silver = (heat gained by vessel) + (heat gained by water)

======> 7308.182718 J = 2113.04 J + (heat absorbed by water)

======> heat absorbed by water = (7308.182718 J) – (2113.04 J)

======> heat absorbed by water = 5195.142718 J = (5195.142718 J)*(1 kJ)/(1000 J)

======> heat absorbed by water = 5.195142718 kJ ≈ 5.195 kJ (ans, correct to 3 sig. figs).