How much heat must be removed from 15.5 g of water at 90.0°C to cool it to 43.2°C?

Heat, Q = 3035.073 J

Explanation:

It is given that,

Mass of water, m = 15.5 g

Initial temperature,

Final temperature,

The specific heat of water is, c = 4.184 J/ g°C

The heat removed or absorbed by water is given by formula as :

So, the heat of 3035.073 J is removed from 15.5 g of water.

*tilts head* sounds more like a statement