Question

An unknown substance has a mass of 0.125 kg and an initial temperature of 95.0°C. The substance is then dropped into a calorimeter made of aluminum containing 0.285 kg of water initially at 25.0°C. The mass of the aluminum container is 0.150 kg, and the temperature of the calorimeter increases to a final equilibrium temperature of 32.0°C. Assuming no thermal energy is transferred to the environment, calculate the specific heat of the unknown substance.

Answer 1

1,186.813J/KgoC

Step-by-step explanation:

Since heat is usually transferred from a hotter to a colder body,

Heat lost by unknown substance = heat gained by aluminum calorimeter + heat gained by water

M(un) x C(un) x [ Temp(un) – Temp(equil) ] = M(Al) x C(Al) x [ Temp(equil) – Temp(Al) ] + M(H2O) x C(H2O) x [ Temp(equil) – Temp(H2O) ]

Where M(un) = 0.125kg, C(un) = specific heat capacity of unknown substance = ?, Temp(un) = 95oC, Temp(equil) = 32oC, M(Al) = 0.150kg, C(Al) = specific heat capacity of aluminum = 921.096J/KgoC, Temp(Al) = 25.0oC, M(H2O) = 0.285Kg, C(H2O)= specific heat capacity of liquid water = 4,200J/KgoC

Temp(H2O) = 25.0oC

That is,

0.125 x C(un) x (95-32) = 0.150 x 921.096 x (32-25) + 0.285 x 4200 x (32-25)

C(un) x 0.125 x 63 = 0.150 x 921.096 x 7 + 0.285 x 4200 x 7

C(un) x 7.875 = 967.1508 + 8379

C(un) x 7.875 = 9346.1508

C(un) = 9346.1508/7.875

C(un) = 1,186.813J/KgoC