The sum of heats gained is zero.
origmasswater*c*(t-30)+newmassadded*c*(t-15)=0
because the density of the rain and pool water is the same, as is the width and length, you can do this wiht depths.
depth originally= 2500/(50*30)
new depth= .005m added.
2500/(50*30)(t-30)+.005(T-15)=0
solve for t
An Olympic swimming pool is heated and maintained at 30 °C in the winter. It has a volume of 2500 m3, a length of 50 m, and a width of 25 m. During a storm, the pool receives 1 cm of rain over its surface area. If the rain is 15 °C, what will the temperature of the pool be after all the water has mixed? Assume the pool heater remains off during this mixing period.
1 answer
1 answer