An electric heating coil is immersed in a beaker of water and turned on. A current of 3.50 A floured through the coil. After the water starts to boil, it is found that 60.0 g of water vaporizes in 4.00 minutes. What is the resistance of the coil? (Iv = 540 cal/g)
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To find the resistance of the coil, we can use the formula:
R = (V^2) / P
Where R is the resistance, V is the voltage across the coil, and P is the power dissipated by the coil.
Since we are not given either the voltage or the power, we need to find one of them in order to calculate the resistance.
We know that the current flowing through the coil is 3.50 A. Using Ohm's Law (V = I * R), we can find the voltage:
V = I * R = 3.50 A * R
Now, we need to find the power dissipated by the coil. Power is given by:
P = E / t
Where P is power, E is the energy, and t is the time.
Given that 60.0 g of water vaporizes in 4.00 minutes, we need to calculate the energy consumed to vaporize this amount of water.
Energy = mass * heat of vaporization.
The heat of vaporization (Iv) is given as 540 cal/g. However, we need to convert calories to joules since power is in watts. We can use the conversion factor: 1 cal = 4.184 J.
So, the heat of vaporization in joules (Iv) is: Iv = 540 cal/g * 4.184 J/cal
Now, calculate the energy consumed:
Energy = mass * Iv
Energy = 60.0 g * Iv
To convert grams to kg, divide the mass by 1000: Energy = 0.0600 kg * Iv
Finally, we can calculate the power:
Power = Energy / t
We need to convert the time from minutes to seconds: t = 4.00 minutes * 60 seconds/minute
Finally, power can be calculated:
Power = 0.0600 kg * Iv / (4.00 minutes * 60 seconds/minute)
Now, we can substitute the values to find the resistance:
R = (V^2) / P
R = (3.50 A * R)^2 / [0.0600 kg * Iv / (4.00 minutes * 60 seconds/minute)]
Simplifying the equation, we can solve for R.