q = mass water x specific heat water x delta T.
You know mass, specific heat water is 4.184 J/g*C and delta T is Tfinal-Tinitial.
A small pebble is heated and placed in a foam cup calorimeter containing 25.0 mL of water at 25.0 degrees Celsius.The water reaches a maximum temperature of 26.4 degrees Celsius. How many joules of heat were released by the pebble?
3 answers
146.3
Volume of Water: 25 mL= 25 g
T1= 25.0 C
T2= 26.4
ΔT= 1.4 C
Q= m CΔT
=25 g x Δ1.8 (26.4-25.0)
=25 x Δ1.8 x1.4= 146.3
T1= 25.0 C
T2= 26.4
ΔT= 1.4 C
Q= m CΔT
=25 g x Δ1.8 (26.4-25.0)
=25 x Δ1.8 x1.4= 146.3
5 answers