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?

25ohms

To find the required resistance of an immersion heater, we can use the formula:

P = (m * c * ΔT) / t

where:
P = power (in watts)
m = mass of water (in kg)
c = specific heat capacity of water (4.186 J/g°C or 4186 J/kg°C)
ΔT = change in temperature (in °C)
t = time (in seconds)

First, we need to convert mass to kilograms and time to seconds:

m = 1.50 kg
t = 10 minutes = 10 * 60 seconds = 600 seconds

Next, we need to calculate the change in temperature:

ΔT = 50°C - 10°C = 40°C

Now, we can substitute the values into the formula:

P = (1.50 kg * 4186 J/kg°C * 40°C) / 600 s

P = 249900 J / 600 s

P ≈ 416.5 W

Since power is equal to voltage multiplied by current (P = V * I), we can rearrange the equation to find the resistance:

R = V^2 / P

R = (120 V)^2 / 416.5 W

R ≈ 34.7 Ω

Therefore, the required resistance of the immersion heater is approximately 34.7 Ω.

To find 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 kilograms,
c is the specific heat capacity of water (approximately 4.18 J/g°C),
ΔT is the change in temperature in degrees Celsius, and
t is the time in seconds.

First, we need to convert the mass of water from kilograms to grams:
1.50 kg = 1500 g

Next, let's convert the time from minutes to seconds:
10 minutes = 10 * 60 = 600 seconds

Now, we can calculate the change in temperature:
ΔT = (50°C - 10°C) = 40°C

Substitute the given values into the formula:
P = (1500 g * 4.18 J/g°C * 40°C) / 600 seconds

Simplify the equation:
P = (1500 g * 4.18 J/g°C * 40°C) / 600 seconds
P = (25,080 J) / 600 seconds
P ≈ 41.8 W

Now, we can use Ohm's Law (V = I * R) to find the resistance of the immersion heater:
V = 120 V (given)

R = V / I

To find the current (I), we can use the formula:
P = V * I

Substitute the given values into the formula:
41.8 W = 120 V * I

Solve for I:
I = 41.8 W / 120 V
I ≈ 0.348 A

Now, we can find the resistance (R):
R = V / I
R = 120 V / 0.348 A
R ≈ 345.98 Ω

Therefore, the required resistance of the immersion heater is approximately 345.98 ohms.

First calculate the heat energy Q needed fr that amount of heating

Q = M C *(delta T) = 1500g*1.00 cal/gm C*40 C = 6*10^4 cal = 2.51*10^5 J

Then require that the electrical power be such that
(V^2/R)*600 s = 2.51*10^5 J

and solve for R