Answer:920.31 J
Explanation:
Given
Volume of water (V)[tex]=14.1 cm^3 [/tex]
mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]
Temperature [tex]=8.4^{\circ} C[/tex]
Final Temperature [tex]=-7.2 ^{\circ}C[/tex]
specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]
Therefore heat required to removed is
[tex]Q=mc(\Delta T)[/tex]
[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]
[tex]Q=920.31 J[/tex]
