Chapter 18, Problem 41P

(a)

To determine

The current drawn from the car’s battery.

Answer to Problem 41P

  $7.18A$

Explanation of Solution

Given:

Specific heat of water, $c_{water}=4186J/kg\cdot °C$

Final temperature, $T_f=95°C$

Initial temperature, $T_i=25°C$

Heater can heat water, $V_{volume}=120mL=120\times {10}^{-3}L=120\times {10}^{-6}{m}^3$

Resistivity of water, ${\rho }_{water}=1000kg/m^3$

Change in time, $\Delta t=8.0\min =8\times 60=480\mathrm{s}$

Voltage, $V=12volts$

Formula used:

By the definition of the power,

  $P=VI$

Where, $V$ is voltage and $I$ is current.

Calculation:

It is known that the efficiency of the heater is about $85\%$ , then the power delivered to the water is about $0.85P_{heater}$ . So,

  $P_{water}=0.85P_{heater}$

It is known that the power of any electric device:

  $P_{heater}=IV$

So, solve this for $I$

  $P_{water}=0.85IV$

  $I=\frac{P_{water}}{0.85V}$

It is also known that the power in general is $P=\frac{\Delta E}{\Delta t}$

Where, $\Delta E$ is the energy transferred to the time $\Delta t$ . Now, calculate the power delivered to the water. For this, find the heat energy transferred to the water by the heater.

Heat energy is $\Delta E=Q=mc\Delta t$ . Where, $c$ is a specific heat of water, $\Delta T$ is the temperature change of water and $m$ is the mass of water. Therefore,

  $P_{water}=\frac{\Delta E}{\Delta T}=\frac{Q_{heat}}{\Delta t}$

  $P_{water}=\frac{Q_{heat}}{\Delta t}$

  $P_{water}=\frac{mc\Delta T}{\Delta t}$

To calculate the mass of $120mL$ water, use the mass density law: $\rho =\frac{m}{V_{volume}}$

So, $m=\rho V_{volume}$

Put this value in expression $P_{water}=\frac{mc\Delta T}{\Delta t}$

  $P_{water}=\frac{\rho V_{volume}c\Delta T}{\Delta t}$

  $I=\frac{P_{water}}{0.85V}=\frac{\frac{\rho V_{volume}c\Delta T}{\Delta t}}{0.85V}$

  $I=\frac{\rho V_{volume}c\Delta T}{0.85V\Delta t}$

  $I=\frac{\rho V_{volume}c\left[T_f-T_i\right]}{0.85V\Delta t}$

Now, substitute all the given values,

  $I=\frac{1000\times 120\times {10}^{-6}\times 4186\times \left[95-25\right]}{0.85\times 12\times 480}=7.18A$

Conclusion: The required current is $7.18A$ .

(b)

To determine

The resistance.

Answer to Problem 41P

  $1.67\Omega$

Explanation of Solution

Given:

Current, $I=7.18A$

Voltage, $V=12volts$

Formula used:

Expression to calculate voltage:

  $V=I\times R$

Where,

  $I$ is the current.

  $V$ is the voltage.

  $R$ is the resistance.

Calculation:

Rearrange the expression in the terms of resistance,

So, $R=\frac{V}{I}$

  $R=\frac{12Volts}{7.18A}=1.67\Omega$

Conclusion: The required resistance is $1.67\Omega$ .