Question
what is the thermal energy change as 175.0g, of water drops from 75.00 dgrs C to 15.50 dgrs C???
q = mass H2O x specific heat water x (Tfinal - Tinitial).
mass H2O = 175.0 g
specific heat water = 4.184 J/g*C
Tf = 15.50
Ti = 75.00
Solve for q.
Check my work for typos.
Thermal energy= mass*specificheat*(Tfinal-Tinitial)
q = mass H2O x specific heat water x (Tfinal - Tinitial).
mass H2O = 175.0 g
specific heat water = 4.184 J/g*C
Tf = 15.50
Ti = 75.00
Solve for q.
Check my work for typos.
Thermal energy= mass*specificheat*(Tfinal-Tinitial)