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Sagot :
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.
Explanation:
The expression for the heat is given as:
[tex]Q=mcT[/tex]Substitute the values in the above equation, we get:
[tex]\begin{gathered} Q=176.251\text{ g}\times4.186\text{ J/g deg C}\times67.714\text{ deg C} \\ \\ Q=49958.4876\text{ J} \end{gathered}[/tex]Joule's equation for electric heating is given as:
[tex]Q=I^2Rt[/tex]Here, R is the resistance of the heating element.
Rearranging the above equation, we get:
[tex]R=\frac{Q}{I^2t}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} R=\frac{49958.4876\text{ J}}{18.014^2\text{ A}^2\times7.167\text{ m}} \\ \\ R=\frac{49958.4876\text{ J}}{324.5042\text{ A}^2\times7.167\times60\text{ s}} \\ \\ R=0.3580\text{ }\Omega \end{gathered}[/tex]Final Answer:
The resistance of the heating element is 0.3580 Ω.
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