change of water temp = +2 deg C
heat energy into water = Cwater*m *2 deg
heat energy out of rock = 532.1 J
so
531 J = Cwater * m * 2
Cwater is specific heat of water in Joules/(gram deg C)
A small pebble is heated and placed in a foam cup calorimeter containing water at 25 °C. The water reaches a maximum temperature of 27 °C. If the pebble released 532.1 J of heat to the water, what mass (in g) of water was in the calorimeter?
can you show me what numbers you plug where, im confused.
5 answers
Specific heat water - 4.187 kJ/kgK
= 4,187 J/gC
= 4,187 J/gC
It would have helped if you had re-posted the question; however, I found it and here is the response I made.
heat lost by pebble + heat gained by water = 0
heat lost by pebble = 532.1 J
heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Set q = heat gained by water and solve for mass H2O.
What's the heat lost by pebble or gained by water. That's 532.1 J.
mass H2O = x
specific heat H2O = 4.194 J/g*C
Tfinal = 27 C
Tinitial = 25 C
So
532.1 = [mass H2O x 4.184 x (27-25)]
heat lost by pebble + heat gained by water = 0
heat lost by pebble = 532.1 J
heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Set q = heat gained by water and solve for mass H2O.
What's the heat lost by pebble or gained by water. That's 532.1 J.
mass H2O = x
specific heat H2O = 4.194 J/g*C
Tfinal = 27 C
Tinitial = 25 C
So
532.1 = [mass H2O x 4.184 x (27-25)]
Specific heat is 4.184 and not 4.194. I typed it right in the equation but wrong above that.
I also typed it wrong 4.184 not 4,184