Equal 0.4 kg masses of lead and tin at 60.0°C are placed in 1.0 kg of water at 20.0°C. cSn = 227 J/ kg ⋅°C ; cPb = 128 J/kg ⋅°C (a) What is the equilibrium temperature of the system? (b) If an alloy is made of half lead and half tin by mass, calculate specific heat of the alloy? (c) How many atoms of tin NSn are in 0.4 kg of tin, and how many atoms of lead NPb are in 0.4 kg of lead? (MSn = 118.7 g/mol ; MPb = 207.2 g/mol) (d)Divide the number NSn of tin atoms by the number NPb of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. What conclusion can be drawn?
Equal 0.4 kg masses of lead and tin at 60.0°C are placed in 1.0 kg of water at 20.0°C. cSn = 227 J/ kg ⋅°C ; cPb = 128 J/kg ⋅°C (a) What is the equilibrium temperature of the system? (b) If an alloy is made of half lead and half tin by mass, calculate specific heat of the alloy? (c) How many atoms of tin NSn are in 0.4 kg of tin, and how many atoms of lead NPb are in 0.4 kg of lead? (MSn = 118.7 g/mol ; MPb = 207.2 g/mol) (d)Divide the number NSn of tin atoms by the number NPb of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. What conclusion can be drawn?
College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter11: Energy In Thermal Processes
Section: Chapter Questions
Problem 23P: Equal 0.400-kg masses of lead and tin at 60.0C are placed in 1.00 kg of water at 20.0C. (a) What is...
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Equal 0.4 kg masses of lead and tin at 60.0°C are placed in 1.0 kg of water at 20.0°C.
cSn = 227 J/ kg ⋅°C ; cPb = 128 J/kg ⋅°C
(a) What is the equilibrium temperature of the system?
(b) If an alloy is made of half lead and half tin by mass, calculate specific heat of the alloy?
(c) How many atoms of tin NSn are in 0.4 kg of tin, and how many atoms of lead NPb are in 0.4 kg of lead? (MSn = 118.7 g/mol ; MPb = 207.2 g/mol)
(d)Divide the number NSn of tin atoms by the number NPb of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. What conclusion can be drawn?
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