Question

a) An immersion heater utilizing\({\rm{120 V}}\)can raise the temperature of a\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ g}}\)aluminium cup containing 350 g of water from to in\({\rm{2}}{\rm{.00 min}}\). Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.

Short answer

1. Resistance of immersion heater for given temperature is\({\rm{14}}{\rm{.8 \Omega }}\).
2. An immersion heater with a lower resistance would shorten the heating time of the water. However, a lower resistance would also increase the current in the wires connected to the heater and there is a limit to the current provided by the power supply.

Step-by-step solution

Definition of power, specific heat of a substance and relation between potential difference & power.

Power: The power of a process is the amount of some type of energy converted into a different type divided by the time interval\({\rm{\Delta t}}\)in which the process occurred:

\({\rm{P = }}\frac{{{\rm{\Delta E}}}}{{{\rm{\Delta t}}}}\) ….. (1)

The SI unit of power is the watt (W).

The energy Q required to change the temperature of a mass m of a substance by an amount\({\rm{\Delta t}}\)is

\({\rm{Q = mc\Delta T}}\) ….. (2)

Where c is the specific heat of the substance.

If a potential difference\({\rm{\Delta V}}\)is maintained across a circuit element, the power, or rate at which energy is supplied to the element, is

\({\rm{P}} = {\rm{I\Delta V}}\) ….. (3)

Because the potential difference across a resistor is given by\({\rm{\Delta V = IR}}\)we can express the power delivered to a resistor as

\({\rm{P = }}{{\rm{I}}^{\rm{2}}}{\rm{R}}\)

\({\rm{P = }}\frac{{{{{\rm{(\Delta V)}}}^{\rm{2}}}}}{{\rm{R}}}\) ….. (4)

The energy delivered to a resistor be electrical transmission appears in the form of internal energy in the resistor.

Given data and required data.

The voltage supplied to the heater,\({\rm{\Delta V = 120\;V}}\)

The mass of the aluminum cup is:

\(\begin{aligned}{}{m_{Al}} &= \left( {1.00 \times {{10}^2}\;g} \right)\left( {\frac{{1\;{\rm{ }}kg}}{{1000{\rm{ }}g}}} \right)\\ &= 0.100{\rm{ }}kg\end{aligned}\)

The mass of the water is:

\(\begin{aligned}{}{m_{water}} &= \left( {350{\rm{ }}g} \right)\left( {\frac{{1\;{\rm{ }}kg}}{{1000{\rm{ }}g}}} \right)\\ &= 0.350{\rm{ }}kg\end{aligned}\)

The temperature difference of the aluminum cup and the water is:

\(\begin{aligned}{}\Delta T &= 95.0^\circ C - 20.0^\circ C\\ &= 75.0^\circ C\end{aligned}\)

The time interval of the heating process is:

\(\begin{aligned}{}\Delta t &= \left( {2.00\;\min } \right)\left( {\frac{{60\;s}}{{1\;\min }}} \right)\\ &= 120{\rm{ }}\;s\end{aligned}\)

The resistance of the heater is constant during the process.

A lower resistance would shorten the heating time.

The specific heat of aluminum is:

\({{\rm{c}}_{{\rm{Al}}}} = {\rm{900\;}}{{\rm{J}} \mathord{\left/ {\vphantom {{\rm{J}} {{\rm{kg}}^\circ {\rm{C}}}}} \right.\\} {{\rm{kg}}^\circ {\rm{C}}}}\)

The specific heat of water is:

\({{\rm{c}}_{{\rm{water }}}} = {\rm{4186\;}}{{\rm{J}} \mathord{\left/ {\vphantom {{\rm{J}} {{\rm{kg}}^\circ {\rm{C}}}}} \right.\\} {{\rm{kg}}^\circ {\rm{C}}}}\)

(a) Calculation for the resistance of heater.

The electrical energy E {electrical supplied by the heater goes into increasing the temperature of the aluminium cup and the} water a temperature difference\({\rm{\Delta T}}\):

\({{\rm{E}}_{{\rm{electrical }}}} = {{\rm{Q}}_{{\rm{Al}}}}{\rm{ + }}{{\rm{Q}}_{{\rm{water }}}}\)

Here,\({{\rm{Q}}_{{\rm{Al}}}}\)is the energy required to raise the temperature of the aluminium cup and\({{\rm{Q}}_{{\rm{water }}}}\)is the energy required to raise the temperature of the water.

Here,\({{\rm{E}}_{{\rm{electrical }}}}\)is found in terms of the power output of the heater and the time interval of the heating process by equation (1)\({{\rm{Q}}_{{\rm{Al}}}}\)and\({{\rm{Q}}_{{\rm{water }}}}\)are found from Equation (2):

\(\begin{aligned}{}{\rm{P\Delta t}} &= {{\rm{m}}_{{\rm{Al}}}}{{\rm{c}}_{{\rm{Al}}}}{\rm{\Delta T + }}{{\rm{m}}_{{\rm{water }}}}{{\rm{c}}_{{\rm{water }}}}{\rm{\Delta TP\Delta t}}\\ &= \left( {{{\rm{m}}_{{\rm{Al}}}}{{\rm{c}}_{{\rm{Al}}}}{\rm{ + }}{{\rm{m}}_{{\rm{water }}}}{{\rm{c}}_{{\rm{water }}}}} \right){\rm{\Delta T}}\end{aligned}\)

Substitute for P from Equation (4):

\(\frac{{{{{\rm{(\Delta V)}}}^{\rm{2}}}}}{{\rm{R}}}{\rm{\Delta t = }}\left( {{{\rm{m}}_{{\rm{Al}}}}{{\rm{c}}_{{\rm{Al}}}}{\rm{ + }}{{\rm{m}}_{{\rm{water }}}}{{\rm{c}}_{{\rm{water }}}}} \right){\rm{\Delta T}}\)

Solve the above equation for R as below.

\(R = \frac{{{{(\Delta V)}^2}\Delta t}}{{\left( {{m_{Al}}{c_{Al}} + {m_{water{\rm{ }}}}{c_{water{\rm{ }}}}} \right)\Delta T}}\)

Substitute numerical values in the above equation.

\(\begin{aligned}{}R &= \frac{{{{\left( {120{\rm{ }}V} \right)}^2} \times 120{\rm{ }}s}}{{\left( {\left( {0.100{\rm{ }}\;kg} \right)\left( {900{\rm{ }}{J \mathord{\left/ {\vphantom {J {kg^\circ C}}} \right.\\} {kg^\circ C}}} \right) + \left( {0.350{\rm{ }}\;kg} \right)\left( {4186{\rm{ }}{J \mathord{\left/ {\vphantom {J {kg^\circ C}}} \right.\\} {kg^\circ C}}} \right)} \right)\left( {75.0^\circ C} \right)}}\\ &= 14.8{\rm{ }}\Omega \end{aligned}\)

Therefore, resistance of immersion heater for given temperature is \(14.8{\rm{ }}\Omega \).

(b) Discuss the practical limits to speeding the heating by lowering the resistance:

An immersion heater with a lower resistance would shorten the heating time of the water. However, a lower resistance would also increase the current in the wires connected to the heater and there is a limit to the current provided by the power supply.