Question

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?

Answer 1

The amount of heat released by the pebble is equal to the amount of heat absorbed by the water, which is given by
$Q=m C_s \Delta T$
where
m is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of the water
$\Delta T = 26.4^{\circ}C-25^{\circ}C=1.4^{\circ}C$ is the increase in temperature of the water.

The density of the water is $1 g/cm^3$, and $1 cm^3=1 mL$, so the mass of the water in the problem is
$m=Vd=(25 mL)(1 g/mL)=25 g$
so if we substitute in the formula, we get the amount of heat absorbed by the water (and released by the pebble):
$Q=m C_s \Delta T=(25 g)(4.18 J/g^{\circ}C)(1.4^{\circ}C)=146.3 J$