Heat Capacity: Difference between revisions
| Line 44: | Line 44: | ||
Solution: Use Q=mCΔT. Q=100 J, m= 3 g, ΔT=3 °C. 100J= (3 g)*(C)*(3 °C). C=11.11111 J/g °C. | Solution: Use Q=mCΔT. Q=100 J, m= 3 g, ΔT=3 °C. 100J= (3 g)*(C)*(3 °C). C=11.11111 J/g °C. | ||
==History== | ==History== | ||
Revision as of 14:52, 31 July 2019
Main Idea
Mathematical Model
Computational Model
Examples
Simple
Middling
Difficult
Connectedness
History
See also
Further reading
External links
References
Heat Capacity
The concept of Heat Capacity is integral to understanding how the temperature of a substance rises and falls. Heat Capacity is the ratio of energy added or removed from a substance to the temperature change observed in that substance. Typically, heat capacities are expressed in terms of the amount of heat (kJ, J, or kCal) that needs to be added to raise the temperature of a substance by 1 degree (Celsius, Fahrenheit, Kelvin). Typical units of Heat Capacities are J/g, kJ/kg, and BTU/lb-mass. The SI unit of heat capacity is J/g.
Calculating/Estimating Heat Capacities
Kopp's Rule
Applications
Examples
Problem: You have a burner that emits 15,000 J of heat in the period that it is left on. Will this burner be able to raise 2 kg of water from 50 °C to 52 °C? The specific heat capacity of water is 4,186 J/kg °C.
Solution: Use Q=mCΔT. The amount of heat needed to do the process specified in the question is Q=(2 kg)*(4,186 J/kg °C)*(2 °C)=16,744 J. Since the burner only gives of 15,000 J, the water will not reach the desired temperature.
Problem: What is the specific heat of 3 g substance that takes 100 J to raise 3 degrees.
Solution: Use Q=mCΔT. Q=100 J, m= 3 g, ΔT=3 °C. 100J= (3 g)*(C)*(3 °C). C=11.11111 J/g °C.
History
See also
Further reading
Elementary Principles of Chemical Processes (3rd Edition) By: Richard M. Felder & Ronald M. Rousseau
Encyclopædia Britannica, 2015, "Heat capacity"