A student reported the following data when calibrating her calorimeter. Calculat

Question

A student reported the following data when calibrating her calorimeter. Calculate its heat capacity(Ccalorimeter) in kJ/•C. Assume the density of water is 1.00g/mL. Initial temp of 50.0mL of warm water=41.0•C Initial temp of 50.0mL of cool water in calorimeter= 24.0•C Final temp of water in calorimeter after adding warm water to calorimeter containing cool water=32.0•C A student reported the following data when calibrating her calorimeter. Calculate its heat capacity(Ccalorimeter) in kJ/•C. Assume the density of water is 1.00g/mL. Initial temp of 50.0mL of warm water=41.0•C Initial temp of 50.0mL of cool water in calorimeter= 24.0•C Final temp of water in calorimeter after adding warm water to calorimeter containing cool water=32.0•C Initial temp of 50.0mL of warm water=41.0•C Initial temp of 50.0mL of cool water in calorimeter= 24.0•C Final temp of water in calorimeter after adding warm water to calorimeter containing cool water=32.0•C

Explanation / Answer

Heat lost by warm water = heat gained by cold water inside calorimeter + heat gained by the calorimeter

q = m * Cp * dt

q = heat, m=mass, Cp=heat capacity and dt = temperature change

so, 50 mL water = 50 g water as the density is 1.00 g /L

50g* 4.184 J/g °C * (41-32)°C = 50g*4.184 J/g °C*(32-24)°C + (calorimeter constant)*(32-24)°C

or, calorimeter constant = 26.15 J/°C = 0.02615 kJ/°C

so, the heat capacity (Ccalorimeter) is 0.02615 kJ/°C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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