Question

To determine the specific heat of an object, a student heats it to 100 ∘C in boiling water. She then places the 87.2 g object in a 173 galuminum calorimeter containing 136 g of water. The aluminum and water are initially at a temperature of 19.8 ∘, and are thermally insulated from their surroundings.

If the final temperature is 23.4 ∘, what is the specific heat of the object?

Referring to the table, identify the material in the object.

Substance	Specific heat, J/(kg⋅K)
Water	4186
Beryllium	1820
Aluminum	900
Glass	837
Silicon	703
Iron (steel)	448
Copper	387
Silver	234
Gold	129
Lead	128

Answer 1

Suppose the Specific Heat capacity of unknown object is C

Heat released/absorbed due to temperature change = Q = m*C*dT

m = mass, C = Specific heat capacity & dT = Tf - Ti

Now Using energy conservation:

Heat gained by water and calorimeter = Heat released by object

Q1 + Q2 = Q3

mw*Cw*dT1 + ma*Ca*dT2 = m*C*dT3

Final temperature of system is 23.4 C

dT1 = Tf - Ti = 23.4 - 19.8 = 3.6 C

dT2 = 23.4 - 19.8 = 3.6 C

dT3 = 100 - 23.4 = 76.6 C

mw = mass of water = 136 gm = 0.136 kg

ma = mass of aluminum = 173 gm = 0.173 kg

m = mass of object = 87.2 gm = 0.0872 kg

Cw = 4186 J/kg-C

Ca = 900 J/kg-C

Now using given values:

C = (mw*Cw*dT1 + ma*Ca*dT2)/(m*dT3)

C = (0.136*4186*3.6 + 0.173*900*3.6)/(0.0872*76.6)

C = 390.7 J/kg-C

So, from above table

Unknown object is Copper.

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