A0.225 kg sample of tin initially at 97.5°c is dropped into 0.115 kg of water. the initial temperature of the water is 10.0°c. if the specific heat capacity of tin is 230 j/kg • °c, what is the final equilibrium temperature of the tin-water mixture

We can take the final temperature to be t
Energy gained by water is given by mcθ , where m is the mass of water, c is the specific heat capacity of water (4186) and θ is change in temperature. 
Energy given by tin sample is equal to the energy gained by water, therefore:
0.115 ×4186 × (t-10) = 0.225 × 230 × (97.5-t)
9.302(t-10) = 97.5 -t
t = (97.5 + 93.02)/10.302 
t = 18.493 degrees Celsius
Therefore, the final temperature is 18.493° C

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