Question

Nonmetric version: (a) How long does a

2.0×105Btu/h2.0 \times 10^5 Btu/h

water heater take to raise the temperature of 65 gal of water from

70F70 ^ { \circ } \mathrm { F }

to

100F100 ^ { \circ } \mathrm { F }

? Metric version: (b) How long does a 59 kW water heater take to raise the temperature of 246 L of water from

21C21 ^ { \circ } \mathrm { C }

to

38F38 ^ { \circ } \mathrm { F }

?

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 3

a)

We can use expression for amount of heat QQ required to change the temperature of a given material and it is proportional to the mass mm of the material present and to the temperature change ΔT\Delta T:

Q=mcwaterΔTcwater=1Btulb°FP=QΔtΔt=QPΔt=mcwaterΔTPρ=mVm=ρV1 gal=231 in31kgm3=3.613105lbin3ρ=1000kgm3=3.613102lbin3Δt=ρVcwaterΔTPΔt=3.613102652311(10070)2105Δt=0.08137 h=4.88min\begin{align*} Q&=mc_{water} \Delta T \\ c_{water}&=1 \, \frac{\text{Btu}}{\text{lb}\cdot \text{\textdegree F}} \tag{specific heat of water from expression 18-15} \\ P&=\frac{Q}{\Delta t} \tag{expression for power} \\ \Delta t&=\frac{Q}{P} \tag{express $\Delta t$} \\ \Delta t&=\frac{mc_{water} \Delta T}{P} \tag{substitute $Q$} \\ \rho&=\frac{m}{V} \tag{expression for density} \\ m&=\rho V \tag{express $m$} \\ 1 \text{ gal}&=231 \text{ in}^3 \tag{from apendix D} \\ 1 \, \frac{\text{kg}}{\text{m}^3}&=3.613 \cdot 10^{-5} \, \frac{\text{lb}}{\text{in}^3} \tag{from apendix D} \\ \rho&=1000\, \frac{\text{kg}}{\text{m}^3}=3.613 \cdot 10^{-2} \, \frac{\text{lb}}{\text{in}^3} \tag{conversion of the density of the water} \\ \Delta t&=\frac{\rho V c_{water} \Delta T}{P} \tag{substitute $m$} \\ \Delta t&=\frac{3.613 \cdot 10^{-2} \cdot 65 \cdot 231 \cdot 1 \cdot (100-70)}{2 \cdot 10^5} \tag{substitute} \\ \Delta t&=0.08137 \text{ h}=\boxed{4.88 \text{min}} \end{align*}

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