This experiment determines the specific heat capacity of a liquid using electrical heating and calorimetry. The electrical energy supplied to the liquid is measured and related to the resulting temperature rise using a graphical method.
When a liquid is heated electrically, the electrical energy supplied is converted into thermal energy. This energy causes an increase in the temperature of the liquid.
The thermal energy gained by a liquid of mass \( m \) and specific heat capacity \( c \) is given by:
\[Q = mc\Delta T\]
The electrical energy supplied by the heater is:
\[E = VIt\]
Assuming negligible heat loss to the surroundings, the electrical energy supplied is equal to the thermal energy gained:
\[VIt = mc\Delta T\]
This can be written in the linear form:
\[E = (mc)\Delta T\]
A graph of electrical energy supplied \( E \) against temperature rise \( \Delta T \) is therefore a straight line whose gradient is equal to \( mc \). The specific heat capacity \( c \) can then be determined.
Record the temperature of the liquid at known time intervals along with the corresponding electrical energy supplied. Calculate the temperature rise \( \Delta T \) from the initial temperature.
For guidance on plotting graphs and determining gradients accurately, visit our Graph Guide.
The graph of electrical energy supplied against temperature rise is a straight line. The specific heat capacity of the liquid is obtained from the gradient of the graph.
Video courtesy of Physics Online (YouTube). Video used for educational purposes under YouTube’s embedding policy.