What is the final temperature after 840 J is absorbed by 10.0 g of water at 25°C?

1 Answer
Dang Thi ThanhHoa · Grace
May 15, 2016

The final temperature is equal to #45^@# #C#

Explanation:

We have the amount of energy gain
#Q=m*c*DeltaT=mcDeltat#

where #c# = #4.184# #J#/#g.C# is the specific heat of water, #m# is the mass of water

#=>##840=10# x #4.184*(t-25)#
#t=840/(10" x "4.184)+25=45# i.e. #45^@# #C#

Related questions
See all questions in Calorimetry
Impact of this question
27301 views around the world
You can reuse this answer
Creative Commons License