College Physics
College Physics
10th Edition
ISBN: 9781285737027
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Textbook Question
Chapter 11, Problem 23P

Equal 0.400-kg masses of lead and tin at 60.0°C are placed in 1.00 kg of water at 20.0°C. (a) What is the equilibrium temperature of the system? (b) If an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy? (c) How many atoms of tin NSn, are in 0.400 kg of tin, and how many atoms of lead NPb are in 0.400 kg of lead? (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 beat of lead. What conclusion can be drawn?

(a)

Expert Solution
Check Mark
To determine
The equilibrium temperature of the system.

Answer to Problem 23P

Solution: The equilibrium temperature of the system is 21.3οC .

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is 60.0οC , mass of water is 1.00 kg and temperature of water is 20.0οC .

Write the expression for equilibrium.

mwcw(TTw)=mcl(TlT)+mct(TlT)

  • m is the mass of lead and tin
  • cl is the specific heat of lead
  • ct is the specific heat of tin
  • mw is the mass of water
  • cw is the specific heat of water
  • T is the equilibrium temperature
  • Tl and Tt are the temperatures of lead and tin respectively
  • Tw is the temperature of water

Re-arrange the above equation to get T.

T=mclTl+mctTt+mwcwTwmcc+mct+mwcw

Substitute 128Jkg1/οC for cl , 218Jkg1/οC for ct , 4186Jkg1/οC for cw , 0.400 kg for m, 1.00 kg for mw , 20οC for Tw , 60οC for Tt and 60οC for Tl in the above equation.

T=(0.400kg)(128Jkg1/οC)(60οC)+(0.400kg)(218Jkg1/οC)(60οC)+(1.00kg)(4186Jkg1/οC)(20οC)(0.400kg)(128Jkg1/οC)+(0.400kg)(218Jkg1/οC)+(1.00kg)(4186Jkg1/οC)=21.28οC21.3οC

Conclusion:

The equilibrium temperature of the system is 21.3οC .

(b)

Expert Solution
Check Mark
To determine
The specific heat of the alloy.

Answer to Problem 23P

Solution: The specific heat of the alloy is 173Jkg1/οC .

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is 60.0οC , mass of water is 1.00 kg and temperature of water is 20.0οC .

The alloy is half lead and half tin by mass. Moreover, the change in temperature of the alloy is same as the change in temperature of lead and tin. Therefore, the specific heat of the alloy is,

ca=cl+ct2

  • ca is the specific heat of the alloy
  • ct is the specific heat of tin
  • cl is the specific heat of lead

Substitute 128Jkg1/οC for cl and 218Jkg1/οC for ct in the above equation.

T=(128Jkg1/οC)+(218Jkg1/οC)2=173Jkg1/οC

Conclusion:

The specific heat of the alloy is 173Jkg1/οC .

(c)

Expert Solution
Check Mark
To determine
The number of atoms of tin and lead.

Answer to Problem 23P

Solution: The number of atoms of tin and lead is 2.03×1024 and 1.16×1024 .

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is 60.0οC , mass of water is 1.00 kg and temperature of water is 20.0οC .

Write the expression to calculate the number of atoms of tin.

Nt=NA(mMt)

  • Nt is the number of atoms of tin
  • NA is the Avogadro number
  • Mt is molar mass of tin

Substitute 6.023×1023mol1 for NA , 0.400 kg for m,and 118.71g/mol for Mt in the above equation to find Nt

Nt=(6.023×1023mol1)(0.400kg118.71g/mol)=(6.023×1023mol1)(0.400×103g118.71g/mol)=2.03×1024

Write the expression to calculate the number of atoms in lead.

Nl=NA(mMl)

  • Nl is the number of atoms of lead
  • NA is the Avogadro number
  • Ml is molar mass of lead

Substitute 6.023×1023mol1 for NA , 0.400 kg for m and 207.2g/mol for Ml to find Nl .

Nl=(6.023×1023mol1)(0.400kg207.2g/mol)=(6.023×1023)(0.400×103g207.2g/mol)=1.16×1024

Conclusion:

Therefore, the number of atoms of tin and lead is 2.03×1024 and 1.16×1024 .

(d)

Expert Solution
Check Mark
To determine
The ratio of number of atoms and specific heat capacities of tin and lead.

Answer to Problem 23P

Solution: The ratio of number of atoms and specific heat capacities of tin and lead is equal.

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is 60.0οC , mass of water is 1.00 kg and temperature of water is 20.0οC .

The ratio of number of atoms of tin and lead is,

r1=NtNl

Substitute 1.93×1024 for Nt and 1.16×1024 for Nl to find rl .

rl=1.93×10241.16×1024=1.66 (I)

The ratio of specific heats of tin and lead is,

r2=ctcl

Substitute 128Jkg1/οC for cl and 218Jkg1/οC for ct to find r2 .

r2=218Jkg1/οC128Jkg1/οC=1.70 (II)

From Equations (I) and (II), the ratio of number of atoms and specific heats of tin and lead is nearly the same.

Conclusion:

The ratio of number of atoms and specific heat capacities of tin and lead is equal

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Chapter 11 Solutions

College Physics

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