Question

A metal object with mass of 20.9 g is heated to 97.0 "Cand then transferred to an insulated container containing 85.1 g of water at 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same temperature of 23.5 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water specific heat: I o e

Answer 1

Amount of heat lost by the metal object = (mass of metal object (m)* specific heat capacity of metal object(s) * change in temperature of the metal object(∆t)

= (20.9 g * s(metal) * (23.5-97.0)°c)= -1536.15 * s(metal)

Heat absorbed by water = - heat lost by metal

Heat absorbed by water= mass of water * specific heat of water * change in temperature of water

= 85.1 g * 4.184 J/g.°c * (23.5-20.5)°c

= 1068.1752 J

Therefore, -1536.15 (g.°c) * specific heat of metal = -1068.1752 J

Specific heat of metal = 1068.1752J/1536.15g.°c

= 0.69536 J/g.°c = 0.69536 J/g.°c * 0.239006 cal/J

= 0.1662 cal/g.°c

Answer: specific heat of metal = 0.1662 cal/g.°c