Question

A 1500-W heater is designed to be plugged into a 120-Voutlet.

What current will flow through the heatingcoil when the heater is plugged in?

I   =    A

What is ,the resistance of the heater?

R =    ohms

How long does it take to raise thetemperature of the air in a good-sized living room $(3.00{\rm m} \times 5.00{\rm m} \times 8.00{\rm m})$ by $10.0^\circ{\rm C}$ ? Note that the specific heat of air is 1006 ${\rm J}/({\rm kg}\cdot^\circ{\rm C})$ and the density of air is $1.20\; {\rm kg}/{\rm m}^3$ .

t=        minutes

Answer 1

Concepts and reason

The concepts required to solve the problem are the current, power and heat energy.

First use the power and the voltage to determine the current and resistance. Then calculate the mass of air using the volume and density. Then use the mass to calculate the total heat energy required to heat up the air. Finally use the energy equation to calculate the time.

Fundamentals

The power is,

$P = IV$

Here, $I$ is the current and $V$ is the voltage.

The power in terms of voltage and resistance is,

$P = \frac{{{V^2}}}{R}$

Here, $R$ is the resistance.

The volume of a rectangular room is,

$V = lbh$

Here, $l$ is the length of the room, $b$ is the width of the room, and $h$ is the height of the room.

The mass is,

$m = \rho V$

Here, $\rho$ is the volume density.

The heat energy is given by

$E = mC\Delta T$

Here, $C$ is the specific heat density and $\Delta T$ is the change in temperature.

The power in terms of energy is,

$P = \frac{E}{t}$

Here, $E$ is the energy and $t$ is the time.

The expression for power is,

$P = IV$

Rearrange the equation in terms of current. The current is,

$I = \frac{P}{V}$

Substitute $1500\,{\rm{V}}$ for $P$ and $120\,{\rm{V}}$ for $V$ in the equation $I = \frac{P}{V}$ to calculate current.

$\begin{array}{c}\\I = \frac{{1500}}{{120}}\\\\ = 12.5\,{\rm{A}}\\\end{array}$

The expression for the power is,

$P = \frac{{{V^2}}}{R}$

Rearrange the equation in terms of resistance. The resistance is,

$R = \frac{{{V^2}}}{P}$

Substitute $1500\,{\rm{V}}$ for $P$ and $120\,{\rm{V}}$ for $V$ in the equation $R = \frac{{{V^2}}}{P}$ to calculate resistance.

$\begin{array}{c}\\R = \frac{{{{\left( {120\,{\rm{V}}} \right)}^2}}}{{1500\,{\rm{W}}}}\\\\ = 9.60\,\Omega \\\end{array}$

The resistance of the heater is $9.60\,\Omega$ .

The volume of the room is,

$V = lbh$

Substitute $5.00\,{\rm{m}}$ for $l$ , $3.00\,{\rm{m}}$ for $b$ , and $8.00\,{\rm{m}}$ for $h$ . The volume of the room is,

$\begin{array}{c}\\V = \left( {5.00\,{\rm{m}}} \right)\left( {3.00\,{\rm{m}}} \right)\left( {8.00\,{\rm{m}}} \right)\\\\ = 120\,{{\rm{m}}^3}\\\end{array}$

The mass of the gas inside the room is,

$m = \rho V$

Substitute $1.20\,{\rm{kg/}}{{\rm{m}}^3}$ for $\rho$ and $120\,{{\rm{m}}^3}$ for $V$ in the above equation $m = \rho V$ to calculate the mass of the gas present in the room.

$\begin{array}{c}\\m = \left( {1.20\,{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {120\,{{\rm{m}}^3}} \right)\\\\ = 144\,{\rm{kg}}\\\end{array}$

The heat energy required to raise the temperature of the air is,

$E = mC\Delta T$

Substitute $144\,{\rm{kg}}$ for $m$ , $1006\,{\rm{J/kg}} \cdot ^\circ {\rm{C}}$ for $C$ , and $10.0^\circ {\rm{C}}$ for $\Delta T$ in the above equation $E = mC\Delta T$ heat energy required to raise the temperature of the air.

$\begin{array}{c}\\E = \left( {144\,{\rm{kg}}} \right)\left( {1006\,{\rm{J/kg}} \cdot {{\rm{m}}^3}} \right)\left( {10.0\,^\circ {\rm{C}}} \right)\\\\ = 1.45 \times {10^6}\,{\rm{J}}\\\end{array}$

The power is,

$P = \frac{E}{t}$

Rewrite the equation in terms of time. The time is,

$t = \frac{E}{P}$

Substitute $1.45 \times {10^6}{\rm{ J}}$ for $E$ and $1500\,{\rm{W}}$ for $P$ in the above equation $t = \frac{E}{P}$ to calculate time required to heat up the room.

$\begin{array}{c}\\t = \frac{{1.45 \times {{10}^6}\,{\rm{J}}}}{{1500\,{\rm{W}}}}\\\\ = 967\,{\rm{s}}\\\\ = {\rm{967}}\,{\rm{s}}\left( {\frac{{1\,{\rm{min}}}}{{60\,{\rm{s}}}}} \right)\\\\ = 16.1\,{\rm{min}}\\\end{array}$

Ans:

The current flow through the heating coil is $12.5\,{\rm{A}}$

The resistance of the heater is $9.60\,\Omega$ .

The time required to heat up the air present in the room is $16.1\,{\rm{minutes}}$ .