A small object of mass m at 100°C is placed into an equal mass of water at 0°C in a calorimeter. The specific heat capacity of the object is half that of the water. Assuming there are no energy transfers to the
environment or to the calorimeter, the final equilibrium temperature of the object plus water will be:
The answer is 33°C, how can i find this answer?
I think this question is lacking some information. Can you recheck it?
For this question you only need to know this :
1- Specific heat capacity formula : q=m.C.DT
I find it useful to re-write the important information given by the question :
1- mass of the X = mass of the water → mx =mH20
2- There is no heat transfer to the environment → this means that all the heat lost by X (the hotter one) will be gained by the water (cooler one) → - qx = qH2O
3- The specific heat capacity of X = 1/2 specific heat capacity of water → Cx = 1/2 CH20
Knowing these you can solve the problem like this :
(click on the image to see the answer if you want)
q = heat (lost or gained)
Cx = specific heat capacity of X
CH2O = specific heat capacity of H2O
Tf = Final temperature