You have 50.0 g of warm water in a calorimeter at 35.0 degree C. You then add 50

Question

You have 50.0 g of warm water in a calorimeter at 35.0 degree C. You then add 50.0 g of cold water at 1.7 degree C and the final temperature of the system was 18.9 degree C. (Specific Heat of Water = 4.184 J (g degree C) Calculate the q for warm water in kJ (should be a negative number) Calculate the q for cold water in kJ (should be a positive number) Using the answers for a. and b., calculate q for the calorimeter (should be negative) Calculate the Heat Capacity for the calorimeter in J/degree C.

Explanation / Answer

a)

q = - m Cp dT

     = - 100 x 4.184 x (35-18.9)

      = -6736 J

     = - 6.74 kJ

b)

q = m Cp dT

q = 100 x 4.184 x (18.9 -1.7)

q = 7196 J

q = 7.20 kJ

c)

q = -q1 - q1

q = + 6.74 - 7.20

q = - 0.46 kJ

d)

Cp = q / dT

      = 0.46 / 18.9

       = 24.3 J / oC

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

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