# Assume that the kinetic energy of a 1600 car moving at 125 is converted entirely into heat. How...

## Question:

Assume that the kinetic energy of a 1600 car moving at 125 is converted entirely into heat. How many calories of heat are released?

## Kinetic Energy

Kinetic energy is the energy possessed by a moving object with a mass. It is dependent on the velocity of the object and the mass of the object. You can compute for the kinetic energy of an object using the formula.

$$\begin{align} E_k = \dfrac{1}{2}mv^2 \end{align} $$

Where:

- {eq}E_k{/eq} is the kinetic energy

**m**is the mass of the object**v**is the velocity of the object

## Answer and Explanation:

Note that for the missing unit we assume that the mass is in kg and the velocity is in {eq}\dfrac{km}{hr}{/eq} since these units are the most appropriate given the nature of the problem.

Given:

- Mass of the car {eq}m = 1600\ \rm kg{/eq}

- Velocity of the car {eq}v = 125\ \rm \dfrac{km}{hr} = 34.7222\ \rm \dfrac{m}{s}{/eq}

Solving for the kinetic energy.

$$\begin{align} E_k &= \dfrac{1}{2}mv^2\\ E_k &= \dfrac{1}{2}(1600\ \rm kg)\left(34.7222\ \rm \dfrac{m}{s}\right)^2\\ E_k &= 964500\ \rm J \end{align} $$

Now we convert it to calories

$$\begin{align} \require{cancel} E_k = 964500\ \rm \cancel{J}\left(\dfrac{1\ \rm cal}{4184\ \rm \cancel{J}}\right) = 230.521033\ \rm cal \end{align} $$

Therefore the calories of heat released is **230.52** cal.

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Chapter 4 / Lesson 14