Chapter 15, Problem 75A

Swimming Pool A swimming pool measuring $20.0m\times 12.5m$ is filled with water to a depth of 3.75m. If the initial temperature is 18.4°C, how much heatmust be added to the water to raise its temperature to29.0°C? Assume that the density of water is 1.000 g/mL.

Interpretation Introduction

Interpretation:

The energy required to heat a given amount of water is to be calculated.

Concept introduction:

Specific heat is defined as quantity of heat required to raise the temperature of one gram substance by one Kelvin. It is represented by ‘c’.

Answer to Problem 75A

The heat absorbed is$4.15\times {10}^{10}J$

Explanation of Solution

The quantity of heat required can be calculated by using formula:

$\begin{array}{l}q=mc\Delta T\\ whereqisquantityofheatabsorbed\\ m=givenmass\\ \Delta T=riseintemperature\\ cisspecificheat\end{array}$

Given data:

$\begin{array}{l}\Delta T=29.0-18.4\\ =10.6°C\\ c=4.184J/g-°C(forwater)\\ Measurementofpool:\\ L=20m\\ B=12.5m\\ Depth=3.75m\\ Densityofwater=1g/mL\end{array}$

$\begin{array}{l}Volume=L\times B\times Depth\\ =20\times 12.5\times 3.75\\ =937.5m^3\\ =937.5\times {10}^{6}mL\end{array}$

$\begin{array}{l}Mass,m=density\times volume\\ =1\times 937.5\times {10}^6\\ =9.375\times {10}^{8}g\end{array}$

Substituting the values in formula:

$\begin{array}{c}q=mc\Delta T\\ =9.375\times {10}^8\times 4.184\times 10.6\\ =4.15\times {10}^{10}J\end{array}$

Conclusion

Thus$4.15\times {10}^{10}J$heat is required to raise the temperature of given amount of water.