Chapter 14, Problem 1OQ

To determine

The direction of heat flow and temperature between two pieces of iron.

Answer to Problem 1OQ

Solution:

B. Heat flows spontaneously from the warm cube to the cold cube until both cubes have same temperature.

Explanation of Solution

Explanation:Given:$cpa$ is the specific heat of iron which is 460 J/kgk

  $cpb$ is the specific heat of iron which is 460 J/kgk

  $ma$ is the mass of warm iron which is 5 kg

  $mb$ is the mass of cold iron which is 10 kg

  $Ta$ is the temperature of warm iron which is $60{}^{\text{ο}}\text{C}$

  $Tb$ is the temperature of cold iron which is $15{}^{\text{ο}}\text{C}$

Formula Used: The formula used is the relation between temperature and specific heat from thermodynamics:

  $Q=mcp(Tf-Ti)$

Where,

  $Q$ is the heat transferred

  $cp$ is the specific heat

  $m$ is the mass

  $Tf$ is the final temperature

  $Ti$ is the initial temperature

From the formula above, the equilibrium temperature formula can be derived.

If two materials $a$ and $b$ with temperatures $Ta$ and $Tb$ are in contact, then they will reach an equilibrium temperature $Te$ .

This equilibrium temperature: $Te$ is given by equating the heat transferred by the two materials.

  $Qa=macpa(Te-Ta)$

  $Qb=mbcpb(Te-Tb)$

Equating the heat transferred is $Qa=-Qb$

This becomes:

  $Qa=macpa(Te-Ta)=-Qb=-mbcpb(Te-Tb)$

  $macpa(Te-Ta)=-mbcpb(Te-Tb)$

  $macpa(Te-Ta)=mbcpb(Tb-Te)$

  $Te=\frac{macpa(Ta)+mbcpb(Tb)}{macpa+mbcpb}$

Where,

  $cpa$ is the specific heat of material $a$

  $cpb$ is the specific heat of material $b$

  $ma$ is the mass of material $a$

  $mb$ is the mass of material $b$

  $Ta$ is the temperature of material $a$

  $Tb$ is the temperature of material $b$

  $Te$ is equilibrium temperature

Calculation: Inserting the given values in the equilibrium temperature formula

  $Te=\frac{macpa(Ta)+mbcpb(Tb)}{macpa+mbcpb}$

  $Te=\frac{(5)(460)(60)+(10)(460)(15)}{(5)(460)+(10)(460)}$

  $Te=45{}^{\text{ο}}\text{C}$

The final equilibrium temperature is 45 C. Since $Te=45{}^{\text{ο}}\text{C}$ is lower that $Ta$ is the temperature of warm iron which is $60{}^{\text{ο}}\text{C}$ and higher that $Tb$ is the temperature of cold iron which is $15{}^{\text{ο}}\text{C}$ , heat flowed from the warm iron to the cold iron. This is consistent with the second law of thermodynamics.

The answer is B.

Conclusion:

Heat flows spontaneously from the warm cube to the cold cube until both cubes have same temperature.