Question

(a) What is the cost of heating a hot tub containing \({\rm{1500 kg}}\) of water from to assuming \({\rm{75}}{\rm{.0\% }}\) efficiency to account for heat transfer to the surroundings? The cost of electricity is \({\rm{12 }}{{{\rm{cents}}} \mathord{\left/ {\vphantom {{{\rm{cents}}} {{\rm{kW}} \cdot {\rm{h}}}}} \right.\\} {{\rm{kW}} \cdot {\rm{h}}}}\).(b) What current was used by the \({\rm{220 - V AC}}\) electric heater, if this took \({\rm{4}}{\rm{.00 h}}\)?

Short answer

1. Cost of the energy supplied to heater is\({\rm{\$ 6}}{\rm{.28}}\).
2. Amount of current used by heater is \({\rm{79}}{\rm{.3\;A}}\).

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)

Here, 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

\({\rm{P = I\Delta V}}\) ….. (3)

Given data and required data:

The mass of water,\({\rm{m}} = {\rm{1500\;kg}}\).

The temperature difference of the water is:

\(\begin{aligned}{}\Delta T &= 40.0^\circ C - 10.0^\circ C\\ &= 30.0^\circ C\end{aligned}\)

The efficiency of the energy conversion is\({\rm{\eta }} = {\rm{75}}{\rm{.0\% }}\).

The cost of electricity is\({\rm{9 }}{{{\rm{cents}}} \mathord{\left/ {\vphantom {{{\rm{cents}}} {{\rm{kW}} \cdot {\rm{h}}}}} \right.\\} {{\rm{kW}} \cdot {\rm{h}}}}\).

The voltage across the electric heater,\({\rm{\Delta V}} = {\rm{220\;V}}\).

The time interval of the heating process is:

\(\begin{aligned}{}{\rm{\Delta t}} &= \left( {{\rm{4}}{\rm{.00\;h}}} \right)\left( {\frac{{{\rm{60\;min}}}}{{{\rm{1\;h}}}}} \right)\left( {\frac{{{\rm{60\;s}}}}{{{\rm{1\;min}}}}} \right)\\ &= {\rm{14400\;s}}\end{aligned}\)

The specific heat of water, \({\rm{c}} = {\rm{4186\;}}{{\rm{J}} \mathord{\left/ {\vphantom {{\rm{J}} {{\rm{kg}} \cdot ^\circ {\rm{C}}}}} \right.\\} {{\rm{kg}} \cdot ^\circ {\rm{C}}}}\)

(a) Calculation of electrical energy supplied by heater and its cost:

The electrical energy\({{\rm{E}}_{{\rm{electrical }}}}\)supplied by the electric heater goes into increasing the temperature the water a temperature difference\({\rm{\Delta T}}\).

But we are given that the efficiency of this energy conversion is\({\rm{75}}{\rm{.0\% }}\).So

\(\left( {{\rm{0}}{\rm{.95}}} \right){{\rm{E}}_{{\rm{clectrical }}}} = {{\rm{Q}}_{{\rm{water}}}}\)

Here,\({{\rm{Q}}_{{\rm{water }}}}\)is the energy required to raise the temperature of the water found from Equation (2):

\(\left( {{\rm{0}}{\rm{.95}}} \right){{\rm{E}}_{{\rm{electrical }}}} = {\rm{mc\Delta T}}\)

Solve the above equation for\({{\rm{E}}_{{\rm{electrical }}}}\)as below.

\({{\rm{E}}_{{\rm{electrical }}}} = \frac{{{\rm{mc\Delta T}}}}{{{\rm{0}}{\rm{.95}}}}\)

Entering the values for m, c, and\({\rm{\Delta T}}\)in the above equation, it gives:

\(\begin{aligned}{}{{\rm{E}}_{{\rm{electrical }}}} &= \frac{{\left( {{\rm{1500\;kg}}} \right)\left( {{\rm{4186\;}}{{\rm{J}} \mathord{\left/ {\vphantom {{\rm{J}} {{\rm{kg}} \cdot ^\circ {\rm{C}}}}} \right.\\} {{\rm{kg}} \cdot ^\circ {\rm{C}}}}} \right)\left( {{\rm{30}}{\rm{.0}}^\circ {\rm{C}}} \right)}}{{{\rm{0}}{\rm{.95}}}}\\ &= {\rm{251}}{\rm{.16 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;J}}\end{aligned}\)

Convert this energy into the units of kilowatt-hours by multiplying by the following conversion factor:

\(\begin{aligned}{}{E_{electrical}} &= \left( {251.16 \times {{10}^6}\;J} \right)\left( {\frac{{1\;{\rm{ }}kW \times h}}{{3.60 \times {{10}^6}\;J}}} \right)\\ &= \frac{{2093}}{{30}}{\rm{ }}\;kW \cdot h\end{aligned}\)

The cost of this energy is found by multiplying by the cost of a\({{\rm{E}}_{{\rm{electrical }}}}\)single kilowatt-hour are given:

\(\begin{aligned}{}Cost{\rm{ }} &= \left( {\frac{{2093}}{{30}}\;kW \cdot h} \right)\left( {\frac{{9{\rm{ }}cents{\rm{ }}}}{{1\;kW \cdot h}}} \right)\\ &= 628{\rm{ }}cents\\ &= (628{\rm{ }}cents{\rm{ }})\left( {\frac{{\$ 1}}{{100{\rm{ }}cents{\rm{ }}}}} \right)\\ &= \$ 6.28\end{aligned}\)

Therefore, Cost of the energy supplied to heater is \({\rm{\$ 6}}{\rm{.28}}\).

(b) Calculation of amount of current used by heater:

The electrical energy\({{\rm{E}}_{{\rm{electrical }}}}\)is found in terms of the power output of the heater and the time interval of the heating process by Equation (2):

\({{\rm{E}}_{{\rm{electrical }}}} = {\rm{P\Delta t}}\)

Substitute for P from equation (3):

\({{\rm{E}}_{{\rm{electrical}}}} = {\rm{I\Delta V\Delta t}}\)

Solve the above equation for I as below.

\({\rm{I = }}\frac{{{{\rm{E}}_{{\rm{electrical }}}}}}{{{\rm{\Delta V\Delta t}}}}\)

Substitute numerical values:

\(\begin{aligned}{}{\rm{I}} &= \frac{{\left( {{\rm{251}}{\rm{.16 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;J}}} \right)}}{{\left( {{\rm{220\;V}}} \right)\left( {{\rm{14400\;s}}} \right)}}\\ &= {\rm{79}}{\rm{.3 A}}\end{aligned}\)

Hence, amount of current used by heater is \({\rm{79}}{\rm{.3\;A}}\).