When boiling water, a hot plate takes an average of 9 minutes and 47 seconds to boil 100 milliliters of water. Assume the temperature in the lab is 75 degrees Fahrenheit. The hot plate is rated to provide 195 watts. How efficient is the hot plate?
Thanks.
To determine the efficiency of the hot plate, we need to calculate the amount of energy it requires to bring the water to boiling point, and then compare it to the amount of energy the hot plate provides.
First, we need to convert the given time to seconds:
9 minutes = 9 x 60 seconds = 540 seconds
47 seconds = 47 seconds
The total time to boil the water is therefore 540 + 47 = 587 seconds.
Next, we need to calculate the energy required to heat the water. We can use the formula:
Energy = mass x specific heat capacity x change in temperature
In this case, the mass of water is 100 milliliters, which is equivalent to 100 grams (since the density of water is 1 gram/cm³).
The specific heat capacity of water is approximately 4.18 J/g°C.
The change in temperature is the final boiling temperature minus the initial room temperature:
Final temperature = 100 degrees Celsius
Initial temperature = 75 degrees Fahrenheit = 23.9 degrees Celsius
Change in temperature = 100 - 23.9 = 76.1 degrees Celsius
Now, let's calculate the energy required:
Energy = 100 g x 4.18 J/g°C x 76.1°C = 31423.8 J
To determine the efficiency, we need to compare the energy provided by the hot plate to the energy required to heat the water.
The power provided by the hot plate is given as 195 watts, and the time taken to boil the water is 587 seconds.
Energy provided = Power x Time = 195 watts x 587 seconds = 114165 watts-second (Joules)
Finally, we can calculate the efficiency by dividing the energy required by the energy provided, and then converting it into a percentage:
Efficiency = (Energy provided / Energy required) x 100
= (114165 J / 31423.8 J) x 100
= 363.35%
So, the efficiency of the hot plate is approximately 363.35%.