A student determined the calorimeter constant of the calorimeter, using the proc

Question

A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 21.65 degree C and of the hot water, 28.75 degree C. The maximum temperature of the mixture was found to be 23.45 degree C. Assume the density of water is 1.00 g mL^-1 and the specific heat is 4.184 J g^-1 K^1. Calculate the heat lost by the hot water. Your Answer:

Explanation / Answer

Heat lost by hot water= mass of hot water* specific heat*temperature difference=

Mass of hot water= Volume* density

50ml*1g/ml*4.18*(28.75-23.45)=1107.7 joules

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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