"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.)"(adsbygoogle = window.adsbygoogle || []).push({});

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 = V^{2}/R

R = V^{2}/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...)

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# Homework Help: What is the resistance of the heater?

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