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Time 30609 90 (s) Temperature (°C) 21.6 27.5 31.3 34.8 38.1 42.4 55.7 0 P= 120 150 180 210 240 Since the energy transferred to the water from the hot plate is E=mcAT, then the power of the hot plate would be: ΔΕ mcAT At At = mc x (slope) -> Calculate the power of your hot plate using m = 0.300 kg, c = 4200 J/kg/°C, and the slope of your graph. . Find a cooled hot plate and read the label on it to see what the manufacturer's power rating is for it.

Time 30609 90 (s) Temperature (°C) 21.6 27.5 31.3 34.8 38.1 42.4 55.7 0 P= 120 150 180 210 240 Since the energy transferred to the water from the hot plate is E=mcAT, then the power of the hot plate would be: ΔΕ mcAT At At = mc x (slope) -> Calculate the power of your hot plate using m = 0.300 kg, c = 4200 J/kg/°C, and the slope of your graph. . Find a cooled hot plate and read the label on it to see what the manufacturer's power rating is for it.

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Just need help solving 2,3 and 4 (Manufacture’s power rating = 460 W) Thank you :)!
Time (s) Temperature (°C) 21.6 27.5 31.3 34.8 38.1 42.4 55.7 030 60 90 120 150 180 210 24 6. Since the energy transferred to the water from the hot plate is E=mcAT, then the power of the hot plate would be: P= ΔΕ mcAT At At efficiency= 4. Why is the efficiency less than 100%? = Calculate the power of your hot plate using m = 0.300 kg, c = 4200 J/kg/°C, and the slope of your graph. 7. Find a cooled hot plate and read the label on it to see what the manufacturer's power rating is for it. Concluding Questions 1. What was your calculated power for the hot plate? 2. Calculate the percent difference between your calculated power and the manufacturer's rating, as follows: manufacturer's rating - calculated power rating manufacturer's power rating % difference= = mc x (slope) 3. Assuming your calculations were correct, and the manufacturer's rating was also correct, what is the ratio of the power you calculated (heat absorbed by the water per second) to the hot plate's rated power (heat given off by the hot plate per second). This ratio is the efficiency of the hot plate. You can convert your decimal fraction to a percent by multiplying by 100. calculated power rating manufacturer's power rating → x 100% x 100%
Transcribed Image Text:Time (s) Temperature (°C) 21.6 27.5 31.3 34.8 38.1 42.4 55.7 030 60 90 120 150 180 210 24 6. Since the energy transferred to the water from the hot plate is E=mcAT, then the power of the hot plate would be: P= ΔΕ mcAT At At efficiency= 4. Why is the efficiency less than 100%? = Calculate the power of your hot plate using m = 0.300 kg, c = 4200 J/kg/°C, and the slope of your graph. 7. Find a cooled hot plate and read the label on it to see what the manufacturer's power rating is for it. Concluding Questions 1. What was your calculated power for the hot plate? 2. Calculate the percent difference between your calculated power and the manufacturer's rating, as follows: manufacturer's rating - calculated power rating manufacturer's power rating % difference= = mc x (slope) 3. Assuming your calculations were correct, and the manufacturer's rating was also correct, what is the ratio of the power you calculated (heat absorbed by the water per second) to the hot plate's rated power (heat given off by the hot plate per second). This ratio is the efficiency of the hot plate. You can convert your decimal fraction to a percent by multiplying by 100. calculated power rating manufacturer's power rating → x 100% x 100%
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