Given 28.2 grams of an unknown substance, if the substance absorbs 2165 joules o

label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

Given 28.2 grams of an unknown substance, if the substance absorbs 2165 joules of energy and the temperature increases by 35 Kelvin, what is the specific heat of the substance?

2.14 x 106 J/g·K
2.19 x 100 J/g·K
5.73 x 10-4 J/g·K
2.69 x 103 J/g·K
Jun 23rd, 2015

Dear your question is under-progress, it would be available to you as soon as it completes within next few minutes.


Jun 23rd, 2015

Specific heat: Q = mcdT. 

Q=heat to change temp of substance = 2165J

m=mass of heated substance= 28.2 g

c=specific heat capacity = ?

dT=temp change = 35K

c = 2165/ (28.2)(35)

c = 2.19 x 100 J/g·K



Jun 23rd, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jun 23rd, 2015
...
Jun 23rd, 2015
Jun 25th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer