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.

"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

obedjefferson8 obedjefferson8 · Answer 2 · 2019-12-24T16:37:41+01:00

Answer:

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