Asked by Richard
A piece of copper block of mass 24g at 230 degree celsius is placed in a copper calorimeter of mass 60g containing 54g of water at 31 degree celsius.Assuming heat losses are negligible,calculate the final steady temperature of the mixture
(Specific heat capacity of water=4200J/kg/k)(Specific heat capacity /of copper=400J/kg/k)
(Specific heat capacity of water=4200J/kg/k)(Specific heat capacity /of copper=400J/kg/k)
Answers
Answered by
Damon
.024 (400) (230 -T) = .060(400)(T-31) + .054(4200)(T-31)
Answered by
Denis
to get the answer
Answered by
faisat
Yes
Answered by
faisat
Faisat
Answered by
Phoebe
Heat lost= 0.024×400(230-T)
Heat gained by water= 0.05×4200(T-31)
Heat gained by calorimeter=0.060×400(T-31)
Heat lost= Heat gained
= 0.024×400(230-T) = 0.54×4200
(T-31)+0.60×400(T-31)
T= 38. 34°C
Heat gained by water= 0.05×4200(T-31)
Heat gained by calorimeter=0.060×400(T-31)
Heat lost= Heat gained
= 0.024×400(230-T) = 0.54×4200
(T-31)+0.60×400(T-31)
T= 38. 34°C
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.