A 11.0 g sample of granite initially at 82.0 ? C is immersed into 23.0 g of wate

Question

A 11.0g sample of granite initially at 82.0?C is immersed into 23.0g of water initially at 25.0?C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, Cs=4.18J/g?C and for granite, Cs=0.790J/g?C.) Express your answer using three significant figures. If someone can please also explain the process in answering the qustion please :( A 11.0g sample of granite initially at 82.0?C is immersed into 23.0g of water initially at 25.0?C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, Cs=4.18J/g?C and for granite, Cs=0.790J/g?C.) Express your answer using three significant figures. If someone can please also explain the process in answering the qustion please :(

Explanation / Answer

heat lost by granite=heat gained by water

(mcdT)for granite = (mcdT) for water

11*0.79*(82-T)=23*4.18*(T-25)

104.83T=3116.08

T=29.725 degree celcius

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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