what is the required resistance of an immersion heater that will increase the temperature of 1.50 kg of water from 10 degrees C to 50 degrees C in 10 minutes while operating 120 V?
heat required=1.5kg*cwater*(50-10) joules
power required= heat/time=heat/600sec
v^2/r= power
resistance=120^2*600/(1.5kj(cwater)40) ohms
check that.
To calculate the required resistance of an immersion heater, we can use the formula:
R = (V^2) / P
where R is the resistance, V is the voltage, and P is the power.
Step 1: Calculate the power required to heat the water.
The amount of heat required to raise the temperature of the water can be calculated using the specific heat capacity formula:
Q = mCΔT
where Q is the heat energy, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m = 1.50 kg (mass of water)
C = 4.18 kJ/(kg·°C) (specific heat capacity of water)
ΔT = 50°C - 10°C = 40°C
Substituting these values into the formula, we have:
Q = (1.50 kg) * (4.18 kJ/(kg·°C)) * 40°C
Q ≈ 250.8 kJ
Since we want to heat the water in 10 minutes, we need to convert the time to seconds:
t = 10 minutes * 60 seconds/minute
t = 600 seconds
The power can be calculated using the formula:
P = Q / t
P = (250.8 kJ) / (600 s)
P ≈ 418 W
Step 2: Calculate the resistance.
We can substitute the given voltage of 120 V and the calculated power of 418 W into the formula mentioned earlier:
R = (V^2) / P
R = (120 V)^2 / 418 W
R = 34.62 Ω
Therefore, the required resistance of the immersion heater to increase the temperature of 1.50 kg of water from 10 degrees C to 50 degrees C in 10 minutes while operating at 120 V is approximately 34.62 Ω.