Question

45g of water at 50C in a beaker is cooled when 50g of copper at 18C is added to it. The contents are stirred till a final constant temperature is reached. calculate this final temperature. the specific heat capacity of copper is 0.39J/g/K and that of water is 4.2J/g/K.

Answer 1

$$\begin{gathered} \mathrm{Heat}\mathrm{gained}\mathrm{by}\mathrm{copper}=\mathrm{m}\times \mathrm{C}\times \mathrm{\theta } \\ =50\times 0.39\times (\mathrm{T}-18) \\ \mathrm{Heat}\mathrm{lost}\mathrm{by}\mathrm{water}=45\times 4.2\times (50-\mathrm{T}) \\ \mathrm{Heat}\mathrm{lost}=\mathrm{Heat}\mathrm{gained} \\ 45\times 4.2\times (50-\mathrm{T})=50\times 0.39\times (\mathrm{T}-18) \\ 9450-189\mathrm{T}=19.5\mathrm{T}-351 \\ 208.5\mathrm{T}=9801 \\ \mathrm{Or}\mathrm{T}={47}^0\mathrm{C}\left(\mathrm{Approx}\right) \\ \mathrm{So}\mathrm{the}\mathrm{fnal}\mathrm{temperature}\mathrm{is}{47}^0\mathrm{C}\left(\mathrm{Approx}\right). \end{gathered}$$