Practical 8 - Measurement of the specific heat capacity for a solid

Question 1

Equation used in this practical

Answer 1

Q = mcΔθ

Question 2

What does Q represent in the equation Q = mcΔθ?

Answer 2

Heat in (J)

Question 3

Way to work out energy using power

Answer 3

Power x time

Question 4

Way to work out energy using current and voltage

Answer 4

Current x voltage x time

Question 5

2 ways in which Q can be worked out in this practical

Answer 5

Power x time
Current x voltage x time

Question 6

Describe the method for this practical using data loggers

Answer 6

1. Connect circuit
2. Measure mass of water in calorimeter
3. Turn on circuit and take sample values of energy supplied and temperature for a range of temperatures
4. Plot a graph of energy (y-axis) vs temperature (x-axis)
5. Gradient x 1/m = c (specific heat capacity)

Question 7

Graph plotted in this practical and how its used to work out the specific heat capacity

Answer 7

Energy (y-axis)
Temperature (x-axis)

Question 8

Traditional (without data loggers) method for this practical

Answer 8

1. Connect circuit
2. Measure mass of water in calorimeter
3. Switch on circuit and start timing. Note down values of voltage and current (which stay constant)
4. Note down temperature at regular time intervals
5. Calculate heat supplied at each time interval (Q=IVt)
6. Plot a graph of energy (y-axis) vs temperature (x-axis)
7. Gradient x 1/m = c (specific heat capacity)

Question 9

How could this experiment be improved? Explain

Answer 9

With better insulation, since there’s a temperature gradient causing heat to escape from the system
For example, a vacuum flask wouldn’t allow any transfer of heat
Also, with water, a constant stirrer would help distribute the heat evenly

Question 10

What type of flask could have been used for this practical and why would this be done?

Answer 10

Vacuum flask
Wouldn’t allow any transfer of heat

Question 11

What would happen the following if the container we used wasn’t a good insulator?
-temperature
-gradient of a graph of temperature v.s time
-specific heat capacity

Answer 11

-all temperature measurements would be lower (heat is lost)
-the gradient of the graph of temperature v.s time would be less steep
-specific heat capacity is larger (overestimated)