Question

For the circuit shown in Fig. P26.58 a 20.0 Ω resistor is embedded in a large block of ice at 0.00°C, and the battery has negligible internal resistance. At what rate (in g/s) is this circuit melting the ice? (The latent heat of fusion for ice is 3.34x10⁵ J/kg.)

Short answer

The circuit will melt ice at $\frac{\mathrm{dm}}{\mathrm{dt}}=6.21\times {10}^{-3}\mathrm{g}/\mathrm{s}$.

Step-by-step solution

Concept Introduction

Power is defined as the energy per unit of time and can be expressed as,

$\mathbf{P}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{R}\mathbf{\hspace{0.33em}}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}$……………………(1)

Given data

$$\begin{gathered} \epsilon =45\hspace{0.33em}V \\ {R}_1=5.00\hspace{0.33em}\Omega \\ {R}_2=10.0\hspace{0.33em}\Omega \\ {R}_3=15.0\hspace{0.33em}\Omega \\ {R}_4=20.0\hspace{0.33em}\Omega \end{gathered}$$

Calculation of current and power

The equivalent resistance of three parallel resistors,

$$\begin{gathered} \frac{1}{{\mathrm{R}}^{\mathrm{\prime }}}=\frac{1}{{\mathrm{R}}_4}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_4} \\ \frac{1}{{\mathrm{R}}^{\mathrm{\prime }}}=\frac{1}{{\mathrm{R}}_4}+\frac{2}{{\mathrm{R}}_4}+\frac{1}{{\mathrm{R}}_4} \\ \frac{1}{{\mathrm{R}}^{\mathrm{\prime }}}=\frac{4}{{\mathrm{R}}_4} \\ {\mathrm{R}}^{\mathrm{\prime }}=\frac{{\mathrm{R}}_4}{4} \end{gathered}$$

Current is the summation of EMFs divided by the summation of resistance,

$$\begin{gathered} I=\frac{\epsilon }{\left(R_1+R_2+R_3+\frac{R_4}{4}\right)} \\ I=\frac{45\mathrm{V}}{5.00\mathrm{\Omega }+10.00\mathrm{\Omega }+15\mathrm{\Omega }+5.0\mathrm{\Omega }} \\ =1.28\mathrm{A} \end{gathered}$$

Use junction rule,

$I=I_1+I_2+I_3$ (2)

V=IR thus,

$$\begin{gathered} I_1{R}_4=I_2{R}_2=I_3{R}_4 \\ 2I_1{R}_2=I_2{R}_2=2I_3{R}_2 \\ 2I_1=I_2=2I_3 \end{gathered}$$

Substituting the values in equation (2), we get

$$\begin{gathered} I=I_1+2I_1+I_1 \\ I=4I_1 \\ {I}_1=\frac{1.28A}{4} \\ {I}_1=0.32A \end{gathered}$$

Calculation of melting rate of ice

The rate of energy in the resistor is,

$$\begin{gathered} P=I^2{R}_4 \\ =\left(0.322\mathrm{A}{)}^2\right(20\mathrm{\Omega }) \\ =2.07\mathrm{W} \end{gathered}$$

is the heat required to melt the ice of mass dm, and L is the latent heat of fusion. thus,

$$\begin{gathered} \mathrm{dE}=\mathrm{Ldm} \\ \frac{\mathrm{dE}}{\mathrm{dt}}=\mathrm{L}\frac{\mathrm{dm}}{\mathrm{dt}} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{P}}{\mathrm{L}} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=\frac{2.07\mathrm{W}}{3.34\times {10}^5\mathrm{J}/\mathrm{kg}} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=6.21\times {10}^{-6}\mathrm{kg}/\mathrm{s}\left(\frac{1000\mathrm{g}/\mathrm{s}}{1\mathrm{kg}/\mathrm{s}}\right) \\ \frac{\mathrm{dm}}{\mathrm{dt}}=6.21\times {10}^{-3}\mathrm{g}/\mathrm{s} \end{gathered}$$

Thus, the circuit will melt ice at $\frac{dm}{dt}=6.21\times {10}^{-3}g/s$.