Question

A particular heating element draws 18.014 A of current and heats 176.251 g of water from 20oC to 87.714oC in 7.167 minutes. What is the resistance of the heating element ? Specific heat of water is 4.186 J/g deg C.

Answer 1

Given:

The current drawn by the heating element is: I = 18.014 A.

Mass of the water is: m = 176.251 g.

The temperature difference of water is: T = 87.714 deg C - 20 deg C = 67.714 deg C.

Time for which the change in temperature occurs: t = 7.167 minutes

The specific heat of water is: c = 4.186 J/g degC.

To find:

The resistance of the heating element.

Step-by-step explanation:

The expression for the heat is given as:

`Q=mcT`

Substitute the values in the above equation, we get:

`\begin{gathered} Q=176.251\text{ g}*4.186\text{ J/g deg C}*67.714\text{ deg C} \\ \\ Q=49958.4876\text{ J} \end{gathered}`

Joule's equation for electric heating is given as:

`Q=I^2Rt`

Here, R is the resistance of the heating element.

Rearranging the above equation, we get:

`R=(Q)/(I^2t)`

Substituting the values in the above equation, we get:

`\begin{gathered} R=\frac{49958.4876\text{ J}}{18.014^2\text{ A}^2*7.167\text{ m}} \\ \\ R=\frac{49958.4876\text{ J}}{324.5042\text{ A}^2*7.167*60\text{ s}} \\ \\ R=0.3580\text{ }\Omega \end{gathered}`

Final Answer:

The resistance of the heating element is 0.3580 Ω.