Study guide: Thermal Models and Gases
Use temperature, heat-capacity and gas models with their limits.
Ideas to explain in your own words
- Kelvin temperature equals Celsius temperature plus 273.15. Absolute-temperature ratios differ from Celsius ratios.
- Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. The model excludes a phase change.
- Final pressure equals initial pressure times final kelvin temperature divided by initial kelvin temperature. Pressure is not predicted by Celsius ratios.
- Heat is energy transfer due to temperature difference, not a material contained in a body. Internal energy is a system property; work can also change it. State the boundary and process.
Useful vocabulary
- K
- Kelvin is the SI thermodynamic temperature unit; its scale has an absolute zero.
- Kelvin temperature
- Kelvin temperature equals Celsius temperature plus 273.15.
- J
- Joule is the SI energy/work unit, equal to one newton-meter.
- Energy transferred
- Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin.
- kPa
- Kilopascal is 1000 pascals; pressure is force per area.
- Final pressure
- Final pressure equals initial pressure times final kelvin temperature divided by initial kelvin temperature.
- temperature
- thermal condition related to average particle motion
Watch your reasoning
- Using the distractor operation a-273.15 instead of a+273.15; Use the 273.15 offset.
- Using the distractor operation a*b+c instead of a*b*c; Use mass in kilograms for this supplied heat-capacity unit.
- Using the distractor operation a*b*c instead of a*b/c; Use kelvin temperatures.
A worked review example
Worked distinct example: A distinct temperature 30°C gives 30+273.15=303.15 K. Method: identify the quantities in the stated rule, substitute using compatible units, and perform the operations in the correct order. Use the 273.15 offset. Kelvin is written without a degree sign. Interpret the final unit and keep the model assumption visible; this calculation is not a measurement of a real learner or site. Worked distinct example: A distinct model has 2 kg, specific heat 150 J/(kg·K) and change 4 K: 2×150×4=1,200 J. Method: identify the quantities in the stated rule, substitute using compatible units, and perform the operations in the correct order. Use mass in kilograms for this supplied heat-capacity unit. A Celsius interval equals the same kelvin interval. Interpret the final unit and keep the model assumption visible; this calculation is not a measurement of a real learner or site. Worked distinct example: A distinct model starts at 100 kPa and 250 K, then reaches 300 K: 100×300/250=120 kPa. Method: identify the quantities in the stated rule, substitute using compatible units, and perform the operations in the correct order. Use kelvin temperatures. Keep gas amount and volume fixed. Interpret the final unit and keep the model assumption visible; this calculation is not a measurement of a real learner or site. Worked supplied problem: A gas gains internal energy by compression while no thermal transfer occurs in the idealized boundary. Explain the pathway. Solution: Work transfers energy into the gas by compression. A change in internal energy does not require heat transfer. Calling internal energy contained heat confuses a system property with a transfer process. This conclusion uses the stated observations and the mechanism explained above; it does not establish every unmeasured feature of a real organism, device or place.
Make a practice plan
- Explain the mechanism in Convert Celsius to kelvin using a labeled case.
- Complete Calculate a heat-capacity model and correct the specific misconception in its feedback.
- Resolve a changed case from Model gas pressure at constant volume, then combine the four ideas in the transfer task.