While transfering the piece of unknown metal to the calorimeter, the student drops the metal into the calorimeter roughly, and some water splashes out of the calorimeter. why must this be avoided? Be specific- how would the loss of water from the calorimeter affect the calculated value of specific heat of the metal?
While transferring the piece of unknown metal to calorimeter, the student drops the metal into thr calorimeter roughly, and some water splashes out of the calorimeter.
As we know,
Heat gained or lost by water
= (mass of water) x (specific heat of water) x deltaT
If some water splashes out it will affect its mass which ultimately affect the calculation of heat.
So this must be avoided.
While calculating the specific heat of metal: -
Specific heat of metal = (mass of water) x (specific heat of water) x deltaT1 / (Mass of metal) x deltaT2
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