What is the resistance of the heater?

[endeavor]

"A student uses an immersion heater to heat a cup of water (300mL) from 20ºC to 80ºC for tea. If it is 75% efficient and takes 2.5 min, what is the resistance of the heater? (Assume 120-V household voltage.)"

Here's what I did:
V= 0.75 * 120 = 90V
mass of water = 0.3 kg, specific heat c = 4186 J/(kg*Cº)
T = change in temp = 60ºC
t = 2.5 min = 150s

Q (heat) = W (work) = mcT = 75348 J
P = W/t = 502.32 W
P = 0.75 * P = 376.74 W
P = V²/R
R = V²/P = 21.5 ohms

First, are my values for voltage and power correct? I wasn't sure where to apply the 75% efficiency.

Second, is my answer correct? it's supposed to be 21 ohms... I changed some values to have exactly 2 significant figures, but my answer just gets farther away from 21. (for example, I tried changing 75348 J to 75000J, but that makes the answer 21.6 ohms, which is farther away from 21 ohms...)

 

[Andrew Mason]

"A student uses an immersion heater to heat a cup of water (300mL) from 20ºC to 80ºC for tea. If it is 75% efficient and takes 2.5 min, what is the resistance of the heater? (Assume 120-V household voltage.)"

Here's what I did:
V= 0.75 * 120 = 90V
mass of water = 0.3 kg, specific heat c = 4186 J/(kg*Cº)
T = change in temp = 60ºC
t = 2.5 min = 150s
...

First, are my values for voltage and power correct? I wasn't sure where to apply the 75% efficiency.

No. Since $P = V^2/R \text{ or } P = I^2R$, the energy consumed by the resistance is proportional to the square of the voltage or current. If it is 75% efficient, then $E = .75 * I^2R\Delta t$

Also, who drinks tea made with 80ºC water? You need to bring it to a boil!

AM

 

[endeavor]

No. Since $P = V^2/R \text{ or } P = I^2R$, the energy consumed by the resistance is proportional to the square of the voltage or current. If it is 75% efficient, then $E = .75 * I^2R\Delta t$

Also, who drinks tea made with 80ºC water? You need to bring it to a boil!

AM

So $$R = .75 * \frac{V^2}{E}\Delta t$$
but Power = E / time, so
$$R = .75 * \frac{V^2}{P}$$
I arrive at the same answer (21.5 ohms) however, but that's probably a coincidence?

 

[Andrew Mason]

Your answer is right actually because you didn't use V=90 volts in your calculation.

AM

 

[thiotimoline]

So $$R = .75 * \frac{V^2}{E}\Delta t$$
but Power = E / time, so
$$R = .75 * \frac{V^2}{P}$$
I arrive at the same answer (21.5 ohms) however, but that's probably a coincidence?

I also got R=21.5 ohms. First, get the total energy needed to boil the water from 20deg to 80deg. Second, take note that energy per 2.5min is only 75% so u need to make it 100%. Finally, apply R =V^2/P to get R.