Question

The heating element of a coffeemaker operates at 120 V and carries a current of 2.00 A. Assuming that the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.500 kg of water from room temperature (23.0°C) to the boiling point. (The specific heat of water is C= 4186 J/Kg oC. The energy required to bring the water to the boiling point is Q = m C ΔT.)

Answer 1

Answer:

$t=671.5s$.

Explanation:

The heat needed will be provided by the power dissipated by the resistor. This power gives certain energy per unit time, so we multiply it by time to get the energy provided: $Q=Pt$.

The equation of heat given mass, specific heat and temperature change is $Q=mC\Delta T$, and the equation of power dissipated by a resistor is $P=Vi$.

Putting all together:

$mC\Delta T=Vit$.

Which means:

$t=\frac{mC\Delta T}{Vi}$.

From 23.0°C to 100°C we have a $\Delta T=77^{\circ}C$, which we won't need to put in Kelvin in this case since it is a variation, so for our values we have:

$t=\frac{(0.5kg)(4186J/kg^{\circ}C)(77^{\circ}C)}{(120V)(2A)}=671.5s$.