Topic Content:

Determination of Specific Heat Capacity of a Solid Using Method of Mixtures

Experiment to Determine the Specific Heat Capacity of a Solid by Method of Mixture:

Aim: Experiment ...

 

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Evaluation Question

Question: Describe an experiment to determine the specific heat capacity of copper using a copper ball.

Answer:

The piece of copper ball is suspended by a thread and placed in a beaker of boiling water.

While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half-filled with water.

The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred quickly to the calorimeter when the water has reached boiling point.

The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the thermometer.

The copper ball is weighed either before or after the experiment.

The result is tabulated as shown below:

Mass of Copper ball = Mcb (kg)

Temperature of Solid = θ

Mass of empty calorimeter + stirrer = M1 kg

Mass of calorimeter + stirrer + water = M2 kg

Mass of water = (M2 – M1) kg

Specific heat capacity of Copper ball = Ccb

Specific heat capacity of water = Cw

Specific heat capacity of material of the calorimeter = Cc

Initial temperature of water & calorimeter =  θ1

Final temperature of mixture =  θ2

Using Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water

Heat lost by copper ball = McbCcb(θ – θ2)

Heat gained by water = M2 – M1 × Cw × (θ2 – θ1)

Heat gained by calorimeter + stirrer = M1 × Cc × (θ2 – θ1) ∴ McbCcb(θ – θ2) = M2 – M1 × Cw × (θ2 – θ1) + M1 × Cc × (θ2 – θ1)

Ccb =  \( \frac {M_2 \:-\: M_1\: \times \: C_w \: \times \: (\theta_2 \:-\: \theta_1) \:+\: M_1\: \times \: C_c \: \times \:(\theta_2 \: – \: \theta_1)}{M_cb(\theta \: – \: \theta_2)} \)

or

Ccb =  \( \frac { \left [(M_2 \:-\: M_1)C_w \:+\: M_1C_c \right ] \left( \theta_2 \: – \: \theta_1 \right)}{M_cb(\theta \: – \: \theta_2)} \)

Precautions

  • The Calorimeter should be lagged. This is done to prevent heat exchange with the surroundings.
  • The hot solid must be transferred quickly from the boiler to the calorimeter.
  • The mixture must be stirred gently to obtain an even temperature.

Evaluation Question

Question:

A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = 4.2Jg−1K−1) (Specific heat capacity of copper = 0.4Jg−1K−1).

Answer:

Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water

McbCcb(θ – θ2) = M2  × Cw × (θ2 – θ1) + M1 × Cc × (θ2 – θ1)

We are to calculate the final temperature θ2

20 × 0.4 × (200 – θ2) = 50 × 4.2 × (θ2 – 30) + 60 × 0.4 × (θ2 – 30)

8(200 – θ2) = 210 (θ2 – 30) + 24 (θ2 – 30)

1600 – 8θ2   = 210θ2 – 6300 + 24θ2 – 720

collect like terms

1600 + 6300 + 720 = 210θ2 + 24θ2 + 8θ2

8620 = 242θ2

θ2 = \(\frac{8620}{242} \)

θ2 = 35.6°C 

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