The amount of heat released by the pebble is equal to the amount of heat absorbed by the water, which is given by
[tex]Q=m C_s \Delta T[/tex]
where
m is the mass of the water
[tex]C_s = 4.18 J/g^{\circ}C[/tex] is the specific heat capacity of the water
[tex]\Delta T = 26.4^{\circ}C-25^{\circ}C=1.4^{\circ}C[/tex] is the increase in temperature of the water.
The density of the water is [tex]1 g/cm^3[/tex], and [tex]1 cm^3=1 mL[/tex], so the mass of the water in the problem is
[tex]m=Vd=(25 mL)(1 g/mL)=25 g[/tex]
so if we substitute in the formula, we get the amount of heat absorbed by the water (and released by the pebble):
[tex]Q=m C_s \Delta T=(25 g)(4.18 J/g^{\circ}C)(1.4^{\circ}C)=146.3 J[/tex]
