3. An experiment was conduc ammo Table 1 that of water. nium nitrate. The heat c

Question

3. An experiment was conduc ammo Table 1 that of water. nium nitrate. The heat capacity of the calorimeter was 132 J/C. Use this information and to answer the following questions. Assume that the specific heat of the solution equals Temperature (ec) Table 1. Heat of Solution Temperature of room Temperature of 100 ml H20 Temperature of ammonium nitrate Lowest temperature after mixing 25.0 25.0 25.0 15.0 A. Calculate the heat released by water (5 pts). (4-104 X 15-25- 6-160 B. Calculate the heat released by the calorimeter (5 pts). C. Calculate the heat of the reaction (5 pts). 4.18+1320-1324

Explanation / Answer

A) Mass of water = Volume x Density = 100 mL x 1 g/mL = 100 g

Q = m x s x (Tf - Ti)

Q = 100 x 4.184 x (15 - 25)

Q = - 4184 J (Heat released by water)

B) Heat released by calorimeter:

Q' = s x (Tf - Ti)

Q' = 132 x ( 15 - 25 )

Q' = - 1320 J

C) Heat of Reaction:

Qt = Q + Q'

Qt = - 4184 - 1320

Qt = - 5504 J

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.