How much heat is added if .617g of water is increased in temperature by .241 degrees C?

1 Answer
Jan 2, 2018

0.622 J

Explanation:

In order to solve this problem, you need the formula:
#q=mC_sDeltaT#

Assuming water is in its liquid state, the specific heat capacity or #C_s# would be 4.18 #J/(g* degrees Celsius)#

#DeltaT# in this case would be 0.241 degrees Celsius since this represents the change in temperature.

#m# would be 0.617 grams.

Plug in all the values and you would get 0.622 joules!
Hope this helps!