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`