The specific heat of lead is 0.129 J/gC. Find the amount of heat released when 2.4 moles of lead are cooled from 37.2C to 22.5C.
Question
Question
The specific heat of lead is 0.129 J/gC. Find the amount of heat released when 2.4 moles of lead are cooled from 37.2C to 22.5C.
Answer:
-943J or 943J released
Explanation:
m=2.4 mol x 207.g/ mol=497.3g
c=.129 J/gC
T= T f-Ti=(22.5-37.2)=-14.7C
Q=?
Q=(49.3g)(.129 J/g C) (-14.7 C)
= -943 J or 943 J released