Asked by Hunter
When 4.62 mL of cold water at 22.6 °C is mixed with 37.5 mL of hot water at 67.9 °C in a thermally isolated calorimeter, what is the final temperature of the mixture when thermal equilibrium has been achieved? H2O density = 1.00 g/mL.
Answers
Answered by
DrBob222
heat lost by warm water + heat gained by cool water = 0
[mass warm water x specific heat water x (Tfinal-Tinitial)] + [mass cool water x specific heat water x (Tfinal-Tinitital)] = 0
Substitute and solve for Tf.
Note that with density H2O of 1.00 g/mL that makes mL H2O the same as grams H2O.
[mass warm water x specific heat water x (Tfinal-Tinitial)] + [mass cool water x specific heat water x (Tfinal-Tinitital)] = 0
Substitute and solve for Tf.
Note that with density H2O of 1.00 g/mL that makes mL H2O the same as grams H2O.
Answered by
Anonymous
[37.5*4.18*(Tf-67.9)+4.62*4.18*(Tf-22.6)
156.75(Tf-67.9)+19.3116(Tf-22.6)
(156.75Tf-10643.325)+(19.3116Tf-436.44216)
176.0616Tf=11079.76716
Tf=11079.76716/176.0616
Tf=62.9
62.9°C
156.75(Tf-67.9)+19.3116(Tf-22.6)
(156.75Tf-10643.325)+(19.3116Tf-436.44216)
176.0616Tf=11079.76716
Tf=11079.76716/176.0616
Tf=62.9
62.9°C
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