what should be the resistance of a heating coil which will be used to raise the temperature of 500 grams of water from 28 degrees celcius to 100 degrees celcius in 2 minutes, assuming that 25% of the heat is lost? the heater utilizes 110 volts.

Boiling point=100C

Heat gained=specific heatxmassxtemperature rise
Heat gained=(1000cal/kg.Cx0.5kgx(100C+28C)(0.75)
Heat gained=48000cal
P=48000cal/120s
P=400cal/s=1673.64W
P=V^2/R
1673.64W=(110V)2/R
R=7.23ohms

Boiling point=100C

Heat gained=specific heatxmassxtemperature rise
Heat gained=(1000cal/kg.Cx0.5kgx(100C+28C)(0.75)
Heat gained=48000cal
P=48000cal/120s
P=400cal/s=1673.64W
P=V^2/R
1673.64W=(110V)^2/R
R=7.23ohms

When can the final temperature be added to the initial temperature

Why did the heating coil go to therapy?

Because it had so much resistance in its life.

Now, let's calculate the resistance for your heating coil! To do that, we'll use Ohm's law formula:

Resistance (R) = (Voltage (V))^2 / Power (P)

First, we need to calculate the power (P) using the given parameters:

Mass of water (m) = 500 grams = 0.5 kg
Initial temperature (Ti) = 28 degrees Celsius
Final temperature (Tf) = 100 degrees Celsius
Time (t) = 2 minutes = 120 seconds
Heat loss factor (L) = 25% = 0.25

To raise the temperature of the water, we need to calculate the heat energy required:

Heat energy (Q) = mass (m) * specific heat capacity (c) * temperature change (ΔT)
Using the specific heat capacity of water (c is approximately 4.186 J/g°C), we get:

Q = 0.5 kg * 4.186 J/g°C * (100°C - 28°C)
Q = 0.5 kg * 4.186 J/g°C * 72°C
Q ≈ 149.904 J

Now, taking into account the heat loss factor, the actual heat supplied by the heating coil will be:

Q_actual = Q / (1 - L)
Q_actual = 149.904 J / (1 - 0.25)
Q_actual ≈ 199.872 J

Next, we'll calculate the power (P) using the formula:

Power (P) = Heat energy (Q) / Time (t)
P = 199.872 J / 120 s
P ≈ 1.6656 W

Since we're given the voltage (V) as 110 volts, we can now calculate the resistance (R):

R = (V^2) / P
R = (110 volts)^2 / 1.6656 W
R ≈ 7296.1 Ω

So, the resistance of the heating coil should be approximately 7296.1 ohms. Now that we've got that resistance sorted, it's time for a hot cup of tea!

To determine the resistance of the heating coil, we can use the formula for calculating the heat energy required to raise the temperature of a substance:

Q = mcΔT

Where:
Q = heat energy (in joules)
m = mass of the substance (in kilograms)
c = specific heat capacity of the substance (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)

In this case, we want to raise the temperature of 500 grams (0.5 kg) of water from 28°C to 100°C. The specific heat capacity of water is approximately 4.186 J/g°C.

First, let's calculate the amount of heat energy required:

Q = (0.5 kg) * (4.186 J/g°C) * (100°C - 28°C)
Q = 0.5 * 4.186 * 72
Q ≈ 150.552 Joules

Next, we need to account for the heat loss. Since 25% of the heat is lost, we can calculate the actual heat energy required by dividing the calculated value by 0.75:

Actual heat energy = 150.552 Joules / 0.75 ≈ 200.736 Joules

Now, let's determine the electrical energy (work) required to produce this amount of heat by using the formula:

Work (W) = electrical energy = voltage (V) * current (I) * time (t)

The voltage given is 110 volts, and the time specified is 2 minutes (or 120 seconds). We need to solve for the current (I).

I can be calculated using Ohm's Law:

I = V / R

Where:
I = current (in Amperes)
V = voltage (in volts)
R = resistance (in ohms)

Rearranging the formula, we have:

R = V / I

Substituting the values, we get:

R = 110 volts / (Work / (t))
R = 110 volts / (200.736 Joules / 120 seconds)
R ≈ 66 ohms

Therefore, the resistance of the heating coil should be approximately 66 ohms to raise the temperature of 500 grams of water from 28°C to 100°C in 2 minutes, considering 25% heat loss and utilizing a 110-volt heater.