Question

A small pebble is heated and placed in a foam cup calorimeter containing 25.0 ml of water at 23.0°c. the water reaches a maximum temperature of 24.2°c. how many joules of heat were released by the pebble? (the specific heat capacity of water is 4.18 j/g•°c and the density of water is 1.00 g/ml)

Answer 1

The increase in temperature of the water is

`\Delta T = 24.2^(\circ)-23.0 ^(\circ)=1.2^(\circ) C`
The total mass of the water is the product between its volume and its density

`m=dV=(1.00 g/mL)(25.0 mL)=25.0 g`
And so we can find the amount of heat released by the pebble to the water, because this is equal to the amount of heat absorbed by the water, which is

`Q=m C_s \Delta T=(25.0 g)(4.18 J/g ^(\circ)C)(1.2 ^(\circ))=125.4 J`