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)
energy in = i^2 R * t
i^2 = 12.25
t = 4*60 = 240 seconds
so energy in = 2940 R Joules
energy to boil = 60 g * 540 cal/g
= 32400 calories
32400 calories(.239J/calorie)
=7744 Joules
so
2940 R = 7744
R = 2.63 Ohms
To find the resistance of the coil, we can use the formula for power:
Power (P) = (Current)^2 * Resistance
We know the current (I) flowing through the coil is 3.50 A.
We also know the power can be defined as the energy used per unit time. In this case, power is the energy supplied to the water in the form of heat, which causes it to vaporize.
We can calculate the energy supplied to the water using the formula:
Energy = Heat Capacity * Mass * Temperature Change
Given:
Mass of water (m) = 60.0 g
Heat capacity of water (Iv) = 540 cal/g
First, let's calculate the energy supplied to the water:
Energy = Heat Capacity * Mass
Energy = 540 cal/g * 60.0 g
Energy = 32,400 cal
Now, we need to convert the energy from calories to joules because the formula for power requires the energy to be in joules:
1 calorie (cal) = 4.18 joules (J)
Energy = 32,400 cal * 4.18 J/cal
Energy = 135,432 J
Since we know the time (t) it took for the water to vaporize is 4.00 minutes, we need to convert it to seconds:
1 minute = 60 seconds
Time (t) = 4.00 minutes * 60 seconds/minute
Time (t) = 240 seconds
Now, we can calculate the power:
Power = Energy / Time
Power = 135,432 J / 240 s
Power = 564.3 J/s (or watts)
Finally, using the power value and the current, we can solve for the resistance:
Power = (Current)^2 * Resistance
564.3 J/s = (3.50 A)^2 * Resistance
Resistance = 564.3 J/s / (3.50 A)^2
Resistance ≈ 45.1 Ω (ohms)
Therefore, the resistance of the coil is approximately 45.1 ohms.