What is the required resistance of an immersion heater that will increase the temperature of 2.00 kg of water from 10.0°C to 50.0°C in 9.0 min while operating at 120 V?
Electrical energy in
= (V^2/R)*time (in Joules)
(use time = 540 seconds)
All of this energy is converted to heat, Q.
The electrical energy required is
Q = M C *(delta T)
= 2000 g*[4.186 J/(deg C*g)]*40 C
Use those two equations to solve for R.
To determine the required resistance of an immersion heater, we can use the formula:
P = (m * c * ΔT) / t
Where:
P is the power in watts
m is the mass of water in kg
c is the specific heat capacity of water (which is approximately 4186 J/kg°C)
ΔT is the change in temperature in °C
t is the time in seconds
First, let's convert the given values:
m = 2.00 kg
ΔT = (50.0°C - 10.0°C) = 40.0°C
t = 9.0 min * 60 sec/min = 540 sec
Now, plug in the values into the formula:
P = (2.00 kg * 4186 J/kg°C * 40.0°C) / 540 sec
P = 626400 J / 540 sec
P ≈ 1160 W
The power is calculated as 1160 watts.
Next, we can use Ohm's Law to find the resistance:
P = V^2 / R
Where:
P is the power in watts
V is the voltage in volts
R is the resistance in ohms
Rearranging the formula to solve for R:
R = V^2 / P
Plugging in the values:
R = (120 V)^2 / 1160 W
R = 14400 V^2 / 1160 W
R ≈ 124.14 Ω
Therefore, the required resistance of the immersion heater is approximately 124.14 ohms.
Electrical energy in
= (V^2/R)*time (in Joules)
(use time = 540 seconds)
All of this energy is converted to heat, Q.
The electrical energy required is
Q = M C *(delta T)
= 2000 g*[4.186 J/(deg C*g)]*40 C
Use those two equations to solve for R.