Calorimetry: final temperature of a mixture (8297)

, por F_y_Q

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100 g of water at 10\ ^oC is mixed with 300 g of water at 40\ ^oC. What will be the final temperature of the mixture?

P.-S.

The heat lost by the hot water is equal to the heat gained by the cold water.

For the hot water: \color[RGB]{2,112,20}{\bm{Q_{hot} =m_{hot}\cdot c_e(H_2O)\cdot (T_f - T_i)}}

For the cold water: \color[RGB]{2,112,20}{\bm{Q_{cold} =m_{cold}\cdot c_e(H_2O)\cdot (T_f - T_i)}}

The heat lost is considered negative, and the heat gained is positive:

\color[RGB]{2,112,20}{\bm{Q_{cold}= - Q_{hot}}}

Equating the heat exchanges:

100\ g\cdot c_e(\ce{H2O})\cdot (T_f - 10\ ^oC) = - 300\ g\cdot c_e(\ce{H2O})\cdot (T_f - 40\ ^oC)

Eliminating the specific heat of water from the equation and simplifying:

-T_f + 10\ ^oC = 3\ T_f - 120\ ^oC

Solving for the final temperature:

130\ ^oC = 4\ T_f\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{32.5\ ^oC = T_f}}}

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