An unknown substance has a mass of 0.125 kg and an initialtemperature of 77.9°C.

Question

An unknown substance has a mass of 0.125 kg and an initialtemperature of 77.9°C. Thesubstance is then dropped into a calorimeter made of aluminumcontaining 0.285 kg of water initially at 30.0°C. The mass of the aluminum container is0.150 kg, and the temperature of the calorimeter increases to afinal equilibrium temperature of 32.0°C. Assuming no thermalenergy is transferred to the environment, calculate the specificheat of the unknown substance.
J/kg·°C

Explanation / Answer

Let the specific heat of the unknown substance be S1. Specific heat of aluminium = S2 = 900 J/kg-C Specific heat of water = S3 = 4184 J/kg-C So, using Q = MSt, => Q1 = 0.125*S1*77.9 + 0.150*900*30 + 0.285*4184*30 Now, final temperature being 32`C, Q2 = [0.125*S1 + 0.150*900 + 0.285*4184]*32 So, equating initial heat Q1 and final heat Q2, S1*125(77.9 - 32) = 0.285*4184*(32-30) + 0.150*900*(32-30) J So, S1 = 462.720 J/kg`C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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