A small pebble is heated and placed in a calorimeter containing 25 ml of water at 25°c. the water reaches a maxiumum temperature of 26.4°c. how many joules of heat were released from the pebble?
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]
