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

Answer:

Heat, Q = 3035.073 J

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 :

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

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