Question

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

Answer 1

Heat, Q = 3035.073 J

Step-by-step explanation:

It is given that,

Mass of water, m = 15.5 g

Initial temperature,
`T_i=90^(\circ) C`

Final temperature,
`T_f=43.2^(\circ)`

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

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

`Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=15.5* 4.184* (43.2-90)\\\\Q=-3035.073\ J`

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