Chapter 27, Problem 30SP

To determine

The resistance of a heating coil which is used to raise the temperature of $500\text{ g}$ of water from $28°\text{C}$ to the boiling point in $2.0\min$, assuming that $25$% of the heat is lost and the heater operates on a $110-\text{V}$ line.

Answer to Problem 30SP

Solution:

The resistance of the heat coil is $7.7\Omega$.

Explanation of Solution

Given data:

Mass of water is $500\text{ g}$.

Initial temperature of water is $28°\text{C}$.

Final temperature of water, that is boiling point of water, is $100°\text{C}$.

Time required to change the temperature is $2\text{ min}$.

The line is of $110\text{ V}$.

Formula used:

The expression of heat is written as,

$Q=mc\left(T_f-T_i\right)$

Here, $Q$ is the heat, $m$ is the mass, $c$ is the specific heat constant $T_i$ is the initial temperature, and $T_f$ is the final temperature.

The expression of energy is written as,

$E=\frac{V^2}{R}t$

Here, $E$ is the energy, $V$ is the voltage, $R$ is the resistance, and $t$ is the time.

Explanation:

The expression to convert initial temperature from degree Celsius to Kelvin is

$T_i=T_i\left(°\text{C}\right)+273.15$

Here, $T_i\left(°\text{C}\right)$ is the temperature in degree Celsius.

Substitute $28°\text{C}$ for $T\left(°\text{C}\right)$

$\begin{array}{c}{T}_i=28°\text{C}+273.15\\ =301.15\text{K}\end{array}$

The expression to convert final temperature from degree Celsius to Kelvin is

$T_f=T_f\left(°\text{C}\right)+273.15$

Here, $T_f\left(°\text{C}\right)$ is the final temperature in degree Celsius.

Substitute $100°\text{C}$ for $T_f\left(°\text{C}\right)$.

$\begin{array}{c}{T}_f=100°\text{C}+273.15\\ =373.15\text{K}\end{array}$

Recall the expression of heat.

$Q=mc\left(T_f-T_i\right)$

Substitute $500\text{ g}$ for $m$, $301.15\text{K}$ for $T_i$, $373.15\text{K}$ for $T_f$, and $4.186\text{ kJ/kg}\cdot \text{K}$ for $c$, that is the specific heat of the water,

$\begin{array}{c}Q=\left(500\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\right)\left(4.186\text{ kJ/kg}\cdot \text{K}\right)\left(373.15\text{K}-301.15\text{K}\right)\\ =150.696\text{ kJ}\end{array}$

Understand that the $25\%$ of heat will be lost. Therefore, total heat that will be required by the heating coil is,

$\begin{array}{c}{Q}_{total}=Q+0.25Q\\ =1.25Q\end{array}$

Here, $Q_{total}$ is the total heat after considering the heat loss.

Substitute $150.696\text{ kJ}$ for $Q$

$\begin{array}{c}{Q}_{total}=1.25\left(150.696\text{ kJ}\right)\\ =188.37\text{ kJ}\end{array}$

Understand that the total heat is equal to the energy dissipated by the heater coil.

Recall the expression of dissipated energy.

$Q_{total}=\frac{V^2}{R}t$

Substitute $188.37\text{ kJ}$ for $Q_{total}$, $110\text{ V}$ for $V$, and $2\text{ min}$ for $t$

$\begin{array}{c}188.37\text{ kJ}\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)=\frac{{\left(110\text{ V}\right)}^2}{R}\left(2\text{ min}\left(\frac{60\text{ s}}{1\text{ min}}\right)\right)\\ R=\frac{\left(12100\right)\left(120\right)}{188370}\\ R=7.7\Omega \end{array}$

Conclusion:

Therefore, the resistance of the required heat coil is $7.7\Omega$.