Answer:
Heat, Q = 3035.073 J
Explanation:
It is given that,
Mass of water, m = 15.5 g
Initial temperature, [tex]T_i=90^{\circ} C[/tex]
Final temperature, [tex]T_f=43.2^{\circ}[/tex]
The specific heat of water is, c = 4.184 J/ g°C
The heat removed or absorbed by water is given by formula as :
[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=15.5\times 4.184\times (43.2-90)\\\\Q=-3035.073\ J[/tex]
So, the heat of 3035.073 J is removed from 15.5 g of water.
