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"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...)
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...)