Life saver rating for complete step by step final answer. Thank you. A copper be

Question

Life saver rating for complete step by step final answer. Thank you.

A copper beaker of mass 0.20 kg contains 0.10 kg of water at 20.0degree C. A block of silver of mass 100 g at 90.0 degree C is added to the beaker of water. The entire system is thermally insulated from the surroundings, (specific heats- copper: 390 J/kg K, silver: 234 J/kg K, water: 4190 J/kg K) What is the final temperature of the system? What is the heat lost by the silver? What is the heat gained by the water? What is the heat gain of the copper?

Explanation / Answer

Mc = 0.20 kg, Tc = 20.0o, Cc = 390 J/kg K

Mw = 0.10 kg, Tw = 20.0o, Cw = 4190 J/kg K

Ms = 0.10 kg, Ts = 90.0o, Cs = 234 J/kg K

a) find T

Ms*Cs(Ts - T) = Mc*Cc(T - Tc) + Mw*Cw(T - Tw)

0.10*234*(90 - T) = 0.20*390*(T - 20) + 0.10*4190*(T - 20)

2106 - 23.4 T = 78 T - 1560 + 419 T - 8380

T = 23.15 oC

b) Ms*Cs(Ts - T) = 1564 J

c) water: Mw*Cw(T - Tw) = 1319 J

copper: Mc*Cc(T - Tc) = 245 J

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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