Question

In the circuit in the figure, a 20-ohm resistor sits inside 104 g of pure water that is surrounded by insulating Styrofoam.
If the water is initially at temperature 11.8 C, how long will it take for its temperature to rise to 57.1 C?
Use 4190 J/kg.C as the heat capacity of water, and express your answer in seconds using three significant figures.

Answer 1

Concepts and reason

The concepts used to solve the problem are Ohm’s law, Joule’s law of heating and theory of heat exchange. The value of the equivalent resistance of the given circuit is first determined using the series and the parallel combination formulae. Then Ohm’s law is applied to the circuit to determine the current flowing in the circuit. The amount of heat given to the water in which one of the resistors is immersed is determined by applying Joule’s law of heating to the resistor. The amount of heat received by water is determined using the specific heat expression. By equating the amount of heat given to the amount of heat received, the duration for which the heat is supplied can be determined.

Fundamentals

Equivalent resistance of a system of resistors is defined as the value of the resistance which when connected across the same source of potential difference as the system of resistors, allows the same current to flow in the circuit.

Equivalent resistance ${R_s}$ of a series circuit is given by the sum of the values of the individual resistances connected in series.

${R_s} = \sum\limits_{i = 1}^n {{R_i}}$ ……(1)

Here, i can take values from 1 to n.

The reciprocal of the equivalent resistance of a parallel circuit is given by the sum of the reciprocals of the individual resistances.

$\frac{1}{{{R_P}}} = \sum\limits_{i = 1}^n {\frac{1}{{{R_i}}}}$ ……(2)

Here, ${R_P}$ is the equivalent resistance of the parallel combination of resistors and i can take values from 1 to n.

When the system of resistors if equivalent resistance R is connected across a battery that applies a potential difference V across the system of resistors, the current I flowing in the circuit is obtained using Ohm’s law. According to Ohm’s law, at constant temperature, the current flowing in a circuit is directly proportional to the potential difference across the circuit. The expression for Ohm’s law can be written as,

$I = \frac{V}{R}$ ……(3)

Current flowing in a resistor generates heat. The value of heat Q generated in a resistor of resistance R, when a current I flows for a time t is given by the Joule’s law of heating. The expression for the heat generated in the resistor is given by,

$Q = {I^2}Rt$ ……(4)

If the heat given by the resistor is absorbed by water of mass m and its temperature rises from its initial temperature ${\theta _i}$ to a final temperature ${\theta _f}$ , then,

$Q = mc\left( {{\theta _f} - {\theta _i}} \right)$ ……(5)

Here, c is the specific heat of water.

If no heat is lost to the surroundings, according to the theory of heat exchange, the heat given by the resistor is equal to the heat absorbed by water.

The time for which the current passes through the resistor is obtained by equating equations (4) and (5).

Draw the circuit diagram representing all the circuit elements as shown below:

Figure 1

The resistors ${R_2}$ and ${R_3}$ are connected in series. Their equivalent resistance ${R_{23}}$ is given by the following expression:

${R_{23}} = {R_2} + {R_3}$ .

Substitute $10.0{\rm{ }}\Omega$ for both ${R_2}$ and ${R_3}$ .

$\begin{array}{c}\\{R_{23}} = {R_2} + {R_3}\\\\ = 10.0{\rm{ }}\Omega + 10.0{\rm{ }}\Omega \\\\ = 20.0{\rm{ }}\Omega \\\end{array}$

In a similar way the equivalent resistance ${R_{45}}$ of the resistors ${R_4}$ and ${R_5}$ connected in series is calculated using the expression,

${R_{45}} = {R_4} + {R_5}$

Substitute $10.0{\rm{ }}\Omega$ for both ${R_4}$ and ${R_5}$ .

$\begin{array}{c}\\{R_{45}} = {R_4} + {R_5}\\\\ = 10.0{\rm{ }}\Omega + 10.0{\rm{ }}\Omega \\\\ = 20.0{\rm{ }}\Omega \\\end{array}$

The equivalent resistance of ${R_6}$ and ${R_7}$ connected in series is ${R_{67}}$ , which is given by the expression

${R_{67}} = {R_6} + {R_7}$

Substitute $5.0{\rm{ }}\Omega$ for both ${R_6}$ and ${R_7}$ .

$\begin{array}{c}\\{R_{67}} = {R_6} + {R_7}\\\\ = 5.0{\rm{ }}\Omega + 5.0{\rm{ }}\Omega \\\\ = 10.0{\rm{ }}\Omega \\\end{array}$

Thus, Fig 1 can be redrawn as follows:

Figure 2

The resistors ${R_{23}}$ , ${R_{45}}$ and ${R_{67}}$ are connected in parallel. If their equivalent resistance is ${R_P}$ then the magnitude of ${R_P}$ is determined using the following equation.

$\frac{1}{{{R_P}}} = \frac{1}{{{R_{23}}}} + \frac{1}{{{R_{45}}}} + \frac{1}{{{R_{67}}}}$

Substitute $20.0{\rm{ }}\Omega$ for both ${R_{23}}$ and ${R_{45}}$ , $10.0{\rm{ }}\Omega$ for ${R_{67}}$ and determine the magnitude of the resistance ${R_P}$ .

$\begin{array}{c}\\\frac{1}{{{R_P}}} = \frac{1}{{{R_{23}}}} + \frac{1}{{{R_{45}}}} + \frac{1}{{{R_{67}}}}\\\\ = \frac{1}{{20.0{\rm{ }}\Omega }} + \frac{1}{{20.0{\rm{ }}\Omega }} + \frac{1}{{10.0{\rm{ }}\Omega }}\\\\ = \frac{1}{{5.0{\rm{ }}\Omega }}\\\end{array}$

Therefore the value of ${R_P}$ is given by,

${R_P} = 5.0{\rm{ }}\Omega$ .

Figure (2) can now be modified as shown below:

Figure 3

The resistances ${R_1}$ , ${R_P}$ and ${R_8}$ are connected in series. The equivalent resistance R of the system of resistors connected in the circuit is given by the following expression:

$R = {R_1} + {R_P} + {R_8}$

Substitute $20.0{\rm{ }}\Omega$ for ${R_1}$ , $5.0{\rm{ }}\Omega$ for both ${R_P}$ and ${R_8}$ . Simplify to obtain R.

$\begin{array}{c}\\R = {R_1} + {R_P} + {R_8}\\\\ = \left( {20.0{\rm{ }}\Omega } \right) + \left( {5.0{\rm{ }}\Omega } \right) + \left( {5.0{\rm{ }}\Omega } \right)\\\\ = 30.0{\rm{ }}\Omega \\\end{array}$

Use equation(3) $I = \frac{V}{R}$ to determine the current flowing in the circuit.

Substitute 30.0 V for V and $30.0{\rm{ }}\Omega$ for R. Simplify for I.

$\begin{array}{c}\\I = \frac{V}{R}\\\\ = \frac{{30.0{\rm{ V}}}}{{30.0{\rm{ }}\Omega }}\\\\ = 1.0{\rm{ A}}\\\end{array}$

The resistor ${R_1}$ is kept immersed in water and current is passed through it. The current passing through ${R_1}$ generates heat, which is transferred to water.

The amount of heat Q generated in the resistor ${R_1}$ due to a current I passing through it for a time t is given by the expression,

$Q = {I^2}{R_1}t$ ……(6)

If no heat is lost to the environment, the heat produced by the resistor is absorbed by water. The amount of heat absorbed by water is given by equation (5) $Q = mc\left( {{\theta _f} - {\theta _i}} \right)$ .

Equate equations (6) and (5) to write an expression for t.

$\begin{array}{c}\\{I^2}{R_1}t = mc\left( {{\theta _f} - {\theta _i}} \right)\\\\t = \frac{{mc\left( {{\theta _f} - {\theta _i}} \right)}}{{{I^2}{R_1}}}\\\end{array}$

Express the mass m of water in kg.

$\begin{array}{c}\\m = 104{\rm{ g}} \times \frac{{{{10}^{ - 3}}{\rm{kg}}}}{{1{\rm{ g}}}}\\\\ = 0.104{\rm{ kg}}\\\end{array}$

Substitute $\left( {0.104{\rm{ kg}}} \right)$ for m, $\left( {4190{\rm{ J/kg}} \cdot ^\circ {\rm{C}}} \right)$ for c, $\left( {57.1^\circ {\rm{C}}} \right)$ for ${\theta _f}$ , $\left( {11.8^\circ {\rm{C}}} \right)$ for ${\theta _i}$ , 1.0 A for I and $\left( {20.0{\rm{ }}\Omega } \right)$ for ${R_1}$ . Solve for t.

$\begin{array}{c}\\t = \frac{{mc\left( {{\theta _f} - {\theta _i}} \right)}}{{{I^2}{R_1}}}\\\\ = \frac{{\left( {0.104{\rm{ kg}}} \right)\left( {4190{\rm{ J/kg}} \cdot ^\circ {\rm{C}}} \right)\left[ {\left( {57.1^\circ {\rm{C}}} \right) - \left( {11.8^\circ {\rm{C}}} \right)} \right]}}{{{{\left( {1.0{\rm{ A}}} \right)}^2}\left( {20.0{\rm{ }}\Omega } \right)}}\\\\ = 987{\rm{ s}}\\\end{array}$

Ans:

The time taken for the water to raise its temperature from 11.8^oC to 57.1^oC is 987 s.