Q. 6 - An Introduction to General, Organic, and Biological Chemistry

Question

A hot tub is filled with $450 gal$of water. $(2.5,2.6,2.7,3.3, 3.4,3.5)$

$\left(a\right)$ What is the volume of water, in liters, in the tub?

$\left(b\right)$ What is the mass, in kilograms, of water in the tub?

$\left(c\right)$ How many kilocalories are needed to heat the water from ${62}^{\circ }F$ to ${105}^{\circ }F$ ?

$\left(d\right)$ If the hot-tub heater provides $5900 kJ/min$, how long, in minutes, will it take to heat the water in the hot tub from ${62}^{\circ }F$ to ${105}^{\circ }F$ ?

 

Step-by-Step Solution

Answer

Part $\left(a\right)$

$\left(a\right)$The volume of water by $V=1.7\times {10}^3\mathrm{L}.$

Part $\left(b\right)$

$\left(b\right)$The mass of water by $M=1.7\times {10}^3 kg.$

Part $\left(c\right)$

$\left(c\right)$The heat need in water as $\text{ Heat }=4.1\times {10}^4\mathrm{kcal}.$

Part $\left(d\right)$

$\left(d\right)$The heat time in tub as $T=0.49h.$

Step: 1 Given information: (Part a)

Given are: $450gal$ of water.

To find the volume of water at tub in liters.

Step: 2 Finding volume of water: (Part a)

The volume of water by 

$\text{ Volume of water in liters }=450\text{ gat }\times \frac{3.785\mathrm{L}}{1\text{ gal }}$

$\begin{array}{l}V=1,703.25\mathrm{L}\\ V=1.7\times {10}^3\mathrm{L}.\end{array}$

As a result,the volume of water in tub at litres is $V=1.7\times {10}^3\mathrm{L}.$

Step: 3 Given information: (Part b)

Given are: $450 gal$ of water.

To find the mass of water interms of kilogram in tub.

Step: 4 Finding mass in tub: (Part b)

The mass of water by 

$Mass of water=1.7\times {10}^{3}L\times \frac{1000\mathrm{mL}}{1L}\times \frac{1.0\mathrm{g}\text{ water }}{1\mathrm{mL}}\times \frac{1\mathrm{kg}}{1000\mathrm{g}}$

$M=1.7\times {10}^3 kg.$

As a result,the mass of water in kilogram is $M=1.7\times {10}^3 kg.$

Step: 5 Given information: (Part c)

Given are: $450 gal$ of water and temperature at ${62}^{\circ }F - {105}^{\circ }F$.

To find the heat .

Step: 6 Finding temperature: (Part c)

The Celsius and Fahrenheit scale are related in the following way:

$T_{\mathrm{c}}=\frac{T_{\mathrm{F}}-32}{1.8}$

Temperature at initial by,

  $T_{\text{initial }}=\frac{{62}^{\circ }\mathrm{F}-32}{1.8}={17}^{\circ }C$

Temperature at final by, 

${\mathrm{T}}_{\text{final }}=\frac{{105}^{\circ }\mathrm{F}-32}{1.8}={41}^{\circ }C.$

Step: 6 Finding heat: (Part c)

Change in temperature,

$\mathrm{\Delta }T=T_{\text{final }}-T_{\text{initial }}$

$\mathrm{\Delta }T={41}^{\circ }\mathrm{C}-{17}^{\circ }\mathrm{C}={24}^{\circ }\mathrm{C}.$

The heat as 

$\text{ Heat }=\text{ mass }\times \mathrm{\Delta }T\times SH$

$\begin{array}{l}\text{ Heat }=1.7\times {10}^3\mathrm{kg}\times {24}^{\circ }\mathrm{C}\times \frac{1.00\mathrm{cal}}{{\mathrm{g}}^{\circ }\mathrm{C}}\times \frac{1\mathrm{kcal}}{{10}^{3}cal}\times \frac{{10}^3\mathrm{g}}{1\mathrm{kg}}\\ \text{ Heat }=40.08\times {10}^3\mathrm{kcal}\\ \text{ Heat }=4.1\times {10}^4\mathrm{kcal}.\end{array}$

Step: 8 Given information: (Part d)

Given are: $4.50 gal$ of water and temperature as ${62}^{\circ }F - {105}^{\circ }F$ and $5900 KJ/min.$

To find time taken to heat in tub.

Step: 9 Heat time in tub: (Part d)

The heat time in tub as 

$T=4.1\times {10}^4\text{ kcal }\times \frac{1\min }{1,400\text{ kcal }}\times \frac{1\mathrm{h}}{60\min }$

$T=0.488h$

$T=0.49h.$

As a result,the heat time in tub is $T=0.49h.$