CP A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at 37.0o C. Suppose that the water were to cool down spontaneously to form ice at 0.0o C, and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20, this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

Solution 96P Step 1 of 4: Pictorial representation of the given problem, Step 2 of 4: Given data, Dimensions of bath tub, a= 190 cm and b= 80 cm Depth of water, c= 24 cm 3 Density of water, = 1 g/cm Mass of the person, m = 70 kg o Initial temperature, T =i7 c o Final temperature, T =f c Change in temperature, T = T T f 0 i7 c = 37 c o Volume of water, V= a × b × c V= 190 ×80×24 3 V = 364800 cm mass of water Density of water, = volume of water Solving for mass of water, M =V 3 3 Substituting V = 364800 cm and = 1 g/cm 3 3 M =(1 g/cm )(364800 cm ) M = 364800 g