The heating element of a coffeemaker operates at 110 V and carries a current of 1.97 A. Assuming that all of the energy transferred from the heating element is absorbed by the water, calculate how long it takes to heat 0.525 kg of water from room temperature (20.1 °C) to the boiling point. Hint: Assume the heat capacity of water to be 4186. J/(kg*C)).
Given mass of water m = 0.525
change in temperature is (T2 - T1) = (100 - 20.1)
heat capacity c = 4186 J/(kg*C)
The amount of heat required for this water is
Q = m * c * (T2-T1)
Q = 0.525 * 4186 * (100-20.1)
Q = 175592.235 J
The heating element supplies the energy at the rate of
power P = V * I
P = 110 * 1.97
P = 216.7 J/s
This is the amount of energy supplid per second
To get an amount of 175592.235 (it takes 't' s) so
216.7 * t = 175592.235
t = 810.3 s
t = 810.3/60 minutes
t = 13.5 min
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