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.

c = 1163.34 J/kg.°C

Explanation:

Specific heat capacity:

"Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass. The specific heat capacity of a material is a physical property."

Use this equation:

mcΔT = (mw c + mAl cAl) ΔT'

Rearranging the equation to find the specific heat (c) you get this:

c = ((mw c + mAl cAl) ΔT') / (mΔT)

c = ((0.285 (4186) + (0.15) (900)) (32 - 25.1)) / ((0.125) (95 - 32))

c = 1163.34 J/kg.°C

1,186.813J/KgoC

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