# A 3.00 watt electric motor is plugged into an electrical outlet. It takes the motor 30.00 seconds...

## Question:

A 3.00 watt electric motor is plugged into an electrical outlet. It takes the motor 30.00 seconds to lift a mass of 254.9 g a distance of 10.00 cm. In that time, the motor has used 90.00 J of energy.

Assuming no energy leaves the system, how much heat has been added to the system by the end of those 30 seconds ?

## Law of Conservation of Energy:

{eq}\\ {/eq}

The Law of Conservation of Energy states that the total energy in an isolated system does not change with time. in simpler terms, energy can neither be created, nor be destroyed, but can only be transformed from one form to the other.

In the given problem, the motor is converting electrical energy into useful work done to raise the mass, and into the generating in the heat the system. The work done by the motor to raise the mass is equal to the change in the gravitational potential energy of the mass.

{eq}\\ {/eq}

We are given:

• the mass lifted by the motor, {eq}m=254.9\;\rm g=0.2459\;\rm kg {/eq}
• The change in the height of the mass, {eq}\Delta h=10\;\rm cm=1.0\;\rm m {/eq}
• The work done by the motor, {eq}W=90\;\rm J {/eq}

The work done by the motor is partially used to raise the mass, and the remaining is converted into heat. The work done to raise a mass at a constant speed is equal to the change in the gravitational potential energy of the mass.

{eq}W=\Delta U+H {/eq}

Here,

• {eq}\Delta U {/eq} is the change in the gravitational potential energy of the mass.
• {eq}H {/eq} is the heat developed.

The change in the gravitational potential energy of an object of mass, {eq}m {/eq}, when its height from the surface of the Earth is changed by {eq}\Delta h {/eq}, is given by the following equation:

{eq}\Delta U=mg\Delta h {/eq}

Here,

• {eq}g=9.8\;\rm m/s^2 {/eq} is the acceleration due to gravity.

Plugging in the given values, we have:

{eq}\begin{align*} \Delta U&=0.2549\times 0.1\times 9.8\\ &=0.2498\;\rm J\\ &\approx 0.25\;\rm J \end{align*} {/eq}

Therefore, the heat developed is:

{eq}\begin{align*} H&=W-\Delta U\\ &=90-0.25\\ &=\boxed{89.75\;\rm J} \end{align*} {/eq} 