Determination of the specific heat capacity of a solid substance by the method of mixtures

When a warm substance and a cold substance are mixed, if no losses of heat to the surroundings take place, then the heat lost by the warm substance is equal to the heat gained by the cold substance. In the experiment referred to above if:

• $m_{1}$: Mass of empty calorimeter and stirrer
• $m_{2}$: Mass of calorimeter with water
• $\theta_{1}$: Initial temperature of the water
• $\theta_{2}$: Temperature of the heated lead shots
• $\theta_{3}$: Maximum temperature of the mixture
• $m_{3}$: Mass of the calorimeter with the mixture
• $c_{1}$: Specific heat capacity of the calorimeter metal
• $c_{2}$: Specific heat capacity of water
• $c_{3}$: Specific heat capacity of lead shots

Then according to the above principle,

Heat lost by lead shots = Heat gained by water + Heat gained by calorimeter

1. Measure the mass $(m_{1})$ of the calorimeter with the stirrer.
2. Fill up to about centimeter below the top of the calorimeter with water at room temperature and measure the mass $(m_{2})$ again.
3. Put a sufficient amount of the substance (lead shots) of which the specific heat capacity ($c_{3}$) is to be found into the tube and heat it with the help of the water heater.
4. After heating till the water boils, measure and record the temperature $(\theta_{2})$ of the lead shots when it becomes constant.
5. Now put the heated lead shots instantly into the water in the calorimeter, stir the mixture well and note down the maximum temperature $(\theta_{3})$ of the mixture. Use the $(0-50)^{\circ}C$ thermometer for this measurement.
6. Finally measure the calorimeter with its contents and note down the mass $(m_{3})$. Use the triple beam balance for all the weighings.
7. Using the standard values for the specific heat capacities of the calorimeter metal and water, calculate the specific heat capacity of lead shots as explained in the theory.