Time remaining:
A student brought 25.50 g of iron ( Cp = 0.449 J/(g°C)) at 23.8°C into contact w

label Chemistry
account_circle Unassigned
schedule 0 Hours
account_balance_wallet $5

A student brought 25.50 g of iron ( Cp = 0.449 J/(g°C)) at 23.8°C into contact with 54.44 g of silver ( Cp = 0.235 J/(g°C)) at 70.5°C. Determine the equilibrium temperature ( to 0.1 °C)

Mar 18th, 2015

Q=mc DT

for iron, q= 25.5*0.449x=11.45x

for silver, q=54.44*0.235x1=12.79x1

q1=q2

12.79/11.45=1.11

(Tf-70.5)/(Tf-23.8)=1.11

Tf-70.5=1.11Tf-26.42

Tf= 44.08= 44.1 degree celcius.......................


Mar 18th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Mar 18th, 2015
...
Mar 18th, 2015
Jun 28th, 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