a) In the experiment Joule's Constant, why should the final temperature be approximately as many...

Question:

a) In the experiment Joule's Constant, why should the final temperature be approximately as many degrees above the room temperature as the initial temperature was below?

b) What is the purpose of measuring the voltage and current at regular intervals during the course of the experiment? Briefly explain the method of determining W.

Joule's Constant:

Joule's constant defines the mechanical work done to increase the temperature of a mass of water by {eq}1^\circ {\rm{C}} {/eq}. It is determined by the ratio of work done to the amount of heat. It is represented by letter 'J.'

Answer and Explanation:


Part (a)

Certain amount of heat energy is involved during the change of temperature of matter.

The expression for heat energy is,

{eq}Q = mC\Delta T {/eq}


From conservation of energy, the amount of work done must be equal to the amount of heat energy produced.

{eq}W = JQ {/eq}


Here, J is Joule?s constant.

{eq}J = \dfrac{W}{Q} {/eq}


Substituting the values in above expression.

{eq}J = \dfrac{{Vq}}{{mC\Delta T}} {/eq}


Expression for current is,

{eq}\begin{align*} i &= \dfrac{q}{t}\\ q &= it \end{align*} {/eq}


Here, q is the quantity of charge and t is the time period.

So, the expression becomes.

{eq}\begin{align*} J &= \dfrac{{Vit}}{{mC\Delta T}}\\ \Delta T &= \dfrac{{Vit}}{{mCJ}} \end{align*} {/eq}


{eq}\Delta T = {T_{final}} - {T_{initial}} {/eq}, is the change in temperature.


Substituting in above expression.

{eq}\begin{align*} \left( {{T_{final}} - {T_{initial}}} \right) &= \dfrac{{Vit}}{{mCJ}}\\ {T_{final}} &= \dfrac{{Vit}}{{mCJ}} + {T_{initial}} \end{align*} {/eq}


Thus, from the above equation the final temperature is greater than the initial temperature. As heating increases the temperature. So, the final temperature be approximately as many degrees above the room temperature as the initial temperature was below.


Part (b)

As, the expression for heat generated per unit time is,

{eq}H = VI {/eq}


So, voltage and current are important to measure for determining the amount of heat generated.


Learn more about this topic:

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