Answer to Question #157819 in General Chemistry for AIM

Question

A swimming pool measuring20.0 m × 12.5 m is filled with water to a depth of 3.75 m. If the initial temperature is 18.4°C, how much heat must be added to the water to raise its temperature to 29.0°C? Assume that the density of water is 1.000 g/mL.

Answer 1

Given:- $l*w = 20*12.5$

Height = 3.75m

$Volume of pool = l*w*h$

volume=20*12*3.75=937.5$m^3$

To calculate mass we have to use this formula

mass=volume*density
density=1g\ml or 10^6 kg/m^3                                      
mass=937.5*10^6gm

we know that
q=mc(T_2-T_1)
q=937.5*10^6*4.18*(29-18.4) [c=4.18j/g^oc]
q=4.16*10^10j