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This lesson focuses on the difference between temperature and heat through an engaging experiment using a Bunsen burner. Students will predict boiling times for 50 cm³ of water and record temperature changes over time. Using a thermometer, they will observe how water’s temperature increases as time progresses, ultimately boiling at 100°C. The experiment will help students understand the relationship between time, temperature, and heat energy, while also allowing them to calculate the power of the Bunsen burner based on observed data.
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Warming up with energy Lesson objectives * know the difference between temperature and heat * Find out through experiment; the power of a bunsen burner
Experiment to compare boiling times for different amounts of water For 50 cm3 water Time / min Temp / C 0.0 18 1.0 50 2.0 80 3.0 92 4.0 100 Prediction: ……………………………………………………….. thermometer 50cm3 water HEAT ……………………………………………………….. Conclusion:
Temp / C 100 80 60 40 20 TIME / mins 0 8 4 2 6
For 50 cm3 water Time / min Temp / C 0.0 18 1.0 50 2.0 80 3.0 92 4.0 100 Temp / C 100 80 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 80 For 50 cm3 water 60 Time / min Temp / C 0.0 18 40 1.0 50 2.0 80 3.0 92 20 4.0 100 TIME / mins 0 8 4 2 6
Temp / C 100 80 Gradient = Δ temp Δ time 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 80 Gradient = Δ temp = Δ T Δ time Δ t 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 80 Gradient = Δ temp = Δ T Δ time Δ t = (68 -33) C 60 sec = 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 80 Gradient = Δ temp = Δ T Δ time Δ t = (68 -33) C 60 sec = 0.58 C / sec 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 Bunsen power: (assuming no energy loss) Δ U = m c Δ T Δ t Δ t = 80 Gradient = Δ temp = Δ T Δ time Δ t = (68 -33) C 60 sec = 0.58 C / sec 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 Bunsen power: (assuming no energy loss) Δ U = m c Δ T Δ t Δ t = 0.05 x 4200 x 0.58 = 80 Gradient = Δ temp = Δ T Δ time Δ t = (68 -33) C 60 sec = 0.58 C / sec 60 40 20 TIME / mins 0 8 4 2 6
Temp / C 100 Bunsen power: (assuming no energy loss) Δ U = m c Δ T Δ t Δ t = 0.05 x 4200 x 0.58 = 121 W 80 Gradient = Δ temp = Δ T Δ time Δ t = (68 -33) C 60 sec = 0.58 C / sec 60 40 20 TIME / mins 0 8 4 2 6