A piece of metal of mass 12 g at 107 ◦c is placed in a calorimeter containing 55.7 g of water at 20◦c. the final temperature of the mixture is 27.6 ◦c. what is the specific heat capacity of the metal? assume that there is no energy lost to the surroundings.

When the heat lost by metal q1 = the heat gained by water q2

and when the heat energy is Q = M*C*ΔT

∴ Mm*Cm*ΔTm = Mw*Cw*ΔTw

when Mm is the mass of metal sample = 12 g

Cm is the specific heat capacity of the metal

ΔTm the difference in temperature in metal = 107-27.6 = 79.4 °C

and:

Mw is the mass of water sample = 55.7 g

Cw is the specific heat capacity of water = 4.18 J. g/°C

ΔTw is the difference in temperature in water = 27.6 - 20 = 7.6 °C

so by substitution, we will get the Cm:

12g * Cm * 79.6 °C = 55.7 g * 4.18 J. g/°C * 7.6°C

∴Cm = (55.7*4.18*7.6) / (12*79.6)

= 1.85 J. g/°C

∴ the specific heat capacity of the metal = 1.85 J. g/°C