Chapter 18, Problem 81QAP

To determine

(a)

The resistance of heater when it is applied $110\text{V}$ due to which $200\text{g}$ of water is heated from $20°\text{C}$ to $90°\text{C}$

Answer to Problem 81QAP

Resistance of heater is $30\Omega$.

Explanation of Solution

Given:

Potential difference across the heater = $V=110\text{V}$

Mass of water = $m=200\text{g}=200\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)=0.2\text{kg}$

Initial temperature of water = $T_1=20°\text{C}$

Final temperature of water = $T_2=90°\text{C}$

Specific heat of water = $c=4186\text{J/kg}°\text{C}$

Time up to heat is on = $t=2.7\min$

Formula used:

Electrical energy consumed in heat is defined as,

  $E=\frac{V^{2}t}{R}$

  $R$ is the resistance.

Heat absorbed by water due to temperature difference is defined as,

  $Q=mc(T_2-T_1)$

Calculation:

It is given that $90\%$ of electrical energy is consumed to heat the water, so,

  $\begin{array}{l}Q=90\%ofE\\ Q=0.9E\\ mc(T_2-T_1)=\frac{0.9V^{2}t}{R}\\ R=\frac{0.9V^{2}t}{mc(T_2-T_1)}\end{array}$

Now, substituting all these values,

  $\begin{array}{l}R=\frac{0.9\times {(110\text{V})}^2\times (162\sec )}{0.2\text{kg}\times 4186{\text{J/kg}}^{\text{o}}\text{C}\times (90°\text{C}-20°\text{C})}=30.1\Omega \\ R=30\Omega \end{array}$

Conclusion:

Thus, resistance of heater is $30\Omega .$

To determine

(b)

The difference in time interval when heater is connected with $12\text{V}$ of battery instead of $110\text{V}$

Answer to Problem 81QAP

The difference in time interval is $3\text{h}45\min$.

Explanation of Solution

Given:

Initial voltage across the heater = $V_1=110\text{V}$

Initial time taken by heater to raise temperature of water with voltage $110\text{V}$,

  $t_1=2.7\min$

Potential difference across the heater when it is connected by car's battery = $V_2=12\text{V}$

Formula used:

Electrical energy consumed in heat is defined as,

  $E=\frac{V^{2}t}{R}$

Calculation:

Since, Electrical energy is defined as,

  $\begin{array}{l}E=\frac{V^{2}t}{R}\\ V^2=\frac{ER}{t}\\ \text{When E and R are constant}\text{. then,}\\ {\text{V}}^2\propto \frac{1}{t}\end{array}$

Consider the time taken by car battery is $t_2$, then,

  $\begin{array}{l}\frac{V_1^{{}^2}}{V_2^2}=\frac{t_2}{t_1}\\ t_2=\frac{V_1^{{}^2}}{V_2^2}\times t_1\\ t_2=\frac{{(110V)}^2}{{(12V)}^2}\times 2.7\min \\ t_2=226.9\min \approx 227\min \end{array}$

Time difference = $t_2-t_1=227\min -2.7\min =224.3min$

  $t_2-t_1=224.3\min \approx 3\text{h}45\min$

Conclusion:

Thus, time difference is $3\text{h}45\min$.