A 0.225 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