Question

(a) What is the cost of heating a hot tub containing $1500\mathrm{k}\mathrm{g}$ of water from to assuming $75.0\mathrm{\%}$ efficiency to account for heat transfer to the surroundings? The cost of electricity is $$\begin{gathered} 12\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}/\phantom{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{k}\mathrm{W}\cdot \mathrm{h}} \\ \mathrm{k}\mathrm{W}\cdot \mathrm{h} \end{gathered}$$.(b) What current was used by the $220-\mathrm{V}\mathrm{A}\mathrm{C}$ electric heater, if this took $4.00\mathrm{h}$?

Short answer

1. Cost of the energy supplied to heater is$\mathrm{\$}6.28$.
2. Amount of current used by heater is $79.3\mathrm{A}$.

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$\mathrm{\Delta }\mathrm{t}$in which the process occurred:

$\mathrm{P}=\frac{\mathrm{\Delta }\mathrm{E}}{\mathrm{\Delta }\mathrm{t}}$ ….. (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$\mathrm{\Delta }\mathrm{T}$is

$\mathrm{Q}=\mathrm{m}\mathrm{c}\mathrm{\Delta }\mathrm{T}$ ….. (2)

Here, c is the specific heat of the substance.

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

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

Given data and required data:

The mass of water,$\mathrm{m}=1500\mathrm{k}\mathrm{g}$.

The temperature difference of the water is:

$\begin{array}{rl}\mathrm{\Delta }T& ={40.0}^{\circ }C-{10.0}^{\circ }C\\ & ={30.0}^{\circ }C\end{array}$

The efficiency of the energy conversion is$\eta =75.0\mathrm{\%}$.

The cost of electricity is$$\begin{gathered} 9\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}/\phantom{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{k}\mathrm{W}\cdot \mathrm{h}} \\ \mathrm{k}\mathrm{W}\cdot \mathrm{h} \end{gathered}$$.

The voltage across the electric heater,$\mathrm{\Delta }\mathrm{V}=220\mathrm{V}$.

The time interval of the heating process is:

$\begin{array}{rl}\mathrm{\Delta }\mathrm{t}& =\left(4.00\mathrm{h}\right)\left(\frac{60\mathrm{m}\mathrm{i}\mathrm{n}}{1\mathrm{h}}\right)\left(\frac{60\mathrm{s}}{1\mathrm{m}\mathrm{i}\mathrm{n}}\right)\\ & =14400\mathrm{s}\end{array}$

The specific heat of water, $$\begin{gathered} \mathrm{c}=4186\mathrm{J}/\phantom{\mathrm{J}\mathrm{k}\mathrm{g}{\cdot }^{\circ }\mathrm{C}} \\ \mathrm{k}\mathrm{g}{\cdot }^{\circ }\mathrm{C} \end{gathered}$$

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

The electrical energy${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$supplied by the electric heater goes into increasing the temperature the water a temperature difference$\mathrm{\Delta }\mathrm{T}$.

But we are given that the efficiency of this energy conversion is$75.0\mathrm{\%}$.So

$\left(0.95\right){\mathrm{E}}_{\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}={\mathrm{Q}}_{\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}$

Here,${\mathrm{Q}}_{\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}$is the energy required to raise the temperature of the water found from Equation (2):

$\left(0.95\right){\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=\mathrm{m}\mathrm{c}\mathrm{\Delta }\mathrm{T}$

Solve the above equation for${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$as below.

${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=\frac{\mathrm{m}\mathrm{c}\mathrm{\Delta }\mathrm{T}}{0.95}$

Entering the values for m, c, and$\mathrm{\Delta }\mathrm{T}$in the above equation, it gives:

$$\begin{gathered} \begin{array}{rl}{\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}& =\frac{\left(1500\mathrm{k}\mathrm{g}\right)\left(4186\mathrm{J}/\phantom{\mathrm{J}\mathrm{k}\mathrm{g}{\cdot }^{\circ }\mathrm{C}} \\ \mathrm{k}\mathrm{g}{\cdot }^{\circ }\mathrm{C}\right)\left(30{.0}^{\circ }\mathrm{C}\right)}{0.95}\\ & =251.16\times 10^6\mathrm{J}\end{array} \end{gathered}$$

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

$\begin{array}{rl}{E}_{electrical}& =\left(251.16\times {10}^{6}J\right)\left(\frac{1kW\times h}{3.60\times {10}^{6}J}\right)\\ & =\frac{2093}{30}kW\cdot h\end{array}$

The cost of this energy is found by multiplying by the cost of a${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$single kilowatt-hour are given:

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

Therefore, Cost of the energy supplied to heater is $\mathrm{\$}6.28$.

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

The electrical energy${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$is found in terms of the power output of the heater and the time interval of the heating process by Equation (2):

${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=\mathrm{P}\mathrm{\Delta }\mathrm{t}$

Substitute for P from equation (3):

${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=\mathrm{I}\mathrm{\Delta }\mathrm{V}\mathrm{\Delta }\mathrm{t}$

Solve the above equation for I as below.

$\mathrm{I}=\frac{{\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}}{\mathrm{\Delta }\mathrm{V}\mathrm{\Delta }\mathrm{t}}$

Substitute numerical values:

$\begin{array}{rl}\mathrm{I}& =\frac{\left(251.16\times 10^6\mathrm{J}\right)}{\left(220\mathrm{V}\right)\left(14400\mathrm{s}\right)}\\ & =79.3\mathrm{A}\end{array}$

Hence, amount of current used by heater is $79.3\mathrm{A}$.