Non-metric version: (a) how long does a $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{Btu}\mathbf{/}\mathbf{h}$water heater take to raise the temperature of data-custom-editor="chemistry" $\mathbf{40}\mathbf{}\mathbf{gal}$ of water from role="math" localid="1662377531189" ${\mathbf{70}}^{\mathbf{\circ }}\mathbf{}\mathbf{F}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{100}}^{\mathbf{\circ }}\mathbf{}\mathbf{F}$? Metric version :(b) How long does a 59 kWwater heater take to raise the temperature of 150 L of water from role="math" localid="1662377622926" ${\mathbf{21}}^{\mathbf{\circ }}\mathbf{C}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{38}}^{\mathbf{\circ }}\mathbf{C}$?

1. The mass of the water is $\mathrm{m}=50\mathrm{gal}\mathrm{or}151.4\mathrm{kg}$
2. The power of heateris$\mathrm{P}=2.0\times {10}^5\frac{\mathrm{Btu}}{\mathrm{h}}\mathrm{or}8.4\mathrm{cal}/\min$
3. The initial temperature of the water is ${\mathrm{T}}_{\mathrm{i}}={70}^{\circ }\mathrm{F}\mathrm{or}{21}^{\circ }\mathrm{C}$
4. The final temperature of the water is ${\mathrm{T}}_{\mathrm{f}}={100}^{\circ }\mathrm{F}\mathrm{or}{38}^{\circ }\mathrm{C}$
5. The power of heater is $\mathrm{P}=59\mathrm{kW}\mathrm{or}59\times {10}^3\mathrm{W}$
6. The volume of the water is $\mathrm{V}=150\mathrm{L}\mathrm{or}150\times {10}^{-3}{\mathrm{m}}^{-3}$
7. The initial temperature of the water is ${\mathrm{T}}_{\mathrm{i}}={21}^{\circ }\mathrm{C}$
8. The final temperature of the water is ${\mathrm{T}}_{\mathrm{f}}={38}^{\circ }\mathrm{C}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the concept of specific heat and the power of the heater. We can convert gallon to kilogram and Btu to cal. The temperature can be converted from Fahrenheit to Celsius.

Formulae:

The heat energy required by a body,$\mathrm{Q}=\mathrm{mc}∆\mathrm{T}$ …(i)

Where, m = mass

c = specific heat capacity

$∆\mathrm{T}$= change in temperature

Q = required heat energy

The power that is exerted by the heat, $\mathrm{P}=\frac{\mathrm{Q}}{\mathrm{t}}$ …(ii)

The specific heat of water,$\mathrm{c}=1000\mathrm{cal}/{\mathrm{kg}}^{\circ }\mathrm{C}$

Comparing equations (i) and (ii), we get that

$\begin{array}{rcl}Pt& =& cm\left(T_f-T_i\right)\\ \mathrm{t}& =& \frac{\mathrm{cm}\left({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}}\right)}{\mathrm{P}}..........................\left(\mathrm{iii}\right)\\ & =& \frac{\left(1000\mathrm{cal}/\mathrm{kg}.{}^{\circ }\mathrm{C}\right)\left(40\mathrm{gal}\right)\left(1000\mathrm{kg}/264\mathrm{gal}\right)\left({100}^{\circ }\mathrm{F}-{70}^{\circ }\mathrm{F}\right)\left(5^{\circ }\mathrm{C}/9^{\circ }\mathrm{F}\right)}{\left(2.0\times {10}^5\mathrm{Btu}/\mathrm{h}\right)\left(252.0\mathrm{cal}/\mathrm{Btu}\right)\left(1\mathrm{h}/60\min \right)}\\ & =& 3.0\min \end{array}$

Hence, the required time to raise the temperature is 3.0 min

Using equation (iii) and the given values, we can get the time required as:

Here,$\mathrm{c}=4190\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$

$\begin{array}{rcl}\mathrm{t}& =& \frac{\left(4190\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}\right)\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(150\mathrm{L}\right)\left(1{\mathrm{m}}^3/1000\mathrm{L}\right)\left({38}^{\circ }\mathrm{C}-{21}^{\circ }\mathrm{C}\right)}{\left(59\times {10}^3\mathrm{J}/\mathrm{s}\right)\times \left(60\mathrm{s}/1\min \right)}\\ & =& 3.0\min \end{array}$

Hence, the time required to raise the temperature is 3.0 min.