Physics
posted by Jimmy .
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.

The sum of heats gained is zero.
origmasswater*c*(t30)+newmassadded*c*(t15)=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)(t30)+.005(T15)=0
solve for t