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Calorimetry: final temperature of a mixture (8297)
Lunes 2 de septiembre de 2024, por
100 g of water at 
is mixed with 300 g of water at 
. What will be the final temperature of the mixture?
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)}}](local/cache-vignettes/L337xH23/7677908c73c18f810c657a3f3848082a-63fcc.png?1733046973 "\color[RGB]{2,112,20}{\bm{Q_{hot} =m_{hot}\cdot c_e(H_2O)\cdot (T_f - T_i)}}"){kind=small}
For the cold water: ![\color[RGB]{2,112,20}{\bm{Q_{cold} =m_{cold}\cdot c_e(H_2O)\cdot (T_f - T_i)}}](local/cache-vignettes/L349xH23/911e7b6c22a10519790749321e41f11e-cd4cf.png?1733046973 "\color[RGB]{2,112,20}{\bm{Q_{cold} =m_{cold}\cdot c_e(H_2O)\cdot (T_f - T_i)}}"){kind=small}
The heat lost is considered negative, and the heat gained is positive:
![\color[RGB]{2,112,20}{\bm{Q_{cold}= - Q_{hot}}}](local/cache-vignettes/L144xH21/473d251aa211eaf398290e1af004d3f5-04364.png?1733046973 "\color[RGB]{2,112,20}{\bm{Q_{cold}= - Q_{hot}}}"){kind=small}
Equating the heat exchanges:
Eliminating the specific heat of water from the equation and simplifying:
Solving for the final temperature:
![130\ ^oC = 4\ T_f\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{32.5\ ^oC = T_f}}}](local/cache-vignettes/L325xH34/2b227ae2c39d10cec30a6a5602010763-656ab.png?1732998278 "130\ ^oC = 4\ T_f\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{32.5\ ^oC = T_f}}}"){kind=small}