Chapter 6, Problem 6.36QP

A 20.0-g block of iron at 50.0^°C and a 20.0 g block of aluminum at 45^°C are placed in contact with each other. Assume that heat is only transferred between the two blocks.

1. a Draw an arrow indicating the heat flow between the blocks.
2. b What is the sign of q_sys for the aluminum when the blocks first come into contact?
3. c What will you observe when q_sys for the iron is zero?
4. d Estimate the temperature of the Al and Fe blocks when q_sys of the iron equals q_sys of the aluminum.

Interpretation Introduction

Interpretation:

For the given process the heat flow direction has to be given.

Answer to Problem 6.36QP

Heat will flow from iron to aluminum block.

Explanation of Solution

Given information,

Mass of the iron block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the iron block is $\text{50}\text{.0}{}^{\text{o}}\text{C}\text{.}$

Mass of the aluminum block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the aluminum block is $\text{45}{}^{\text{o}}\text{C}\text{.}$

Heat will flow from iron to aluminum block.

$\text{Iron}\text{block }\left(\text{50}{}^{\text{o}}\text{C}\right)\stackrel{\left(\text{heat flow direction}\right)}{\to }\text{Aluminum}\text{block }\left(\text{45}{}^{\text{o}}\text{C}\right)$

Interpretation Introduction

Interpretation:

The sign of ${\text{q}}_{\text{system}}$ for the aluminum block has to be given.

Answer to Problem 6.36QP

The sign of ${\text{q}}_{\text{system}}$ for the aluminum block is positive

Explanation of Solution

Given information,

Mass of the iron block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the iron block is $\text{50}\text{.0}{}^{\text{o}}\text{C}\text{.}$

Mass of the aluminum block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the aluminum block is $\text{45}{}^{\text{o}}\text{C}\text{.}$

The colder aluminum block absorbing heat from the hotter iron block.  Hence, the sign of the ${\text{q}}_{\text{system}}$ of the aluminum block is positive $\left({\text{q}}_{\text{system}}\text{>}\text{0}\right)\text{.}$

Interpretation Introduction

Interpretation:

The observation when ${\text{q}}_{\text{system}}$ for the iron is zero has to be given.

Answer to Problem 6.36QP

Both aluminum block and the iron block will have the same temperature.

Explanation of Solution

Given information,

Mass of the iron block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the iron block is $\text{50}\text{.0}{}^{\text{o}}\text{C}\text{.}$

Mass of the aluminum block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the aluminum block is $\text{45}{}^{\text{o}}\text{C}\text{.}$

The temperature of the both blocks will be the same when ${\text{q}}_{\text{system}}$ of iron block is zero.

$\text{Iron}\text{block}{\text{q}}_{\text{sys}}\text{=}\text{Aluminum}{\text{block q}}_{\text{sys}}$

Interpretation Introduction

Interpretation:

The temperature of $\text{Al}\text{and}\text{Fe}$ blocks has to be given.

Answer to Problem 6.36QP

The temperature of the $\text{Al}\text{and}\text{Fe}$ blocks will be $\text{46}\text{.7}{}^{\text{o}}\text{C}\text{.}$

Explanation of Solution

Given information,

Mass of the iron block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the iron block is $\text{50}\text{.0}{}^{\text{o}}\text{C}\text{.}$

Mass of the aluminum block $\text{20}\text{.0}\text{g}\text{.}$  temperature of the aluminum block is $\text{45}{}^{\text{o}}\text{C}\text{.}$

The temperature calculation of both blocks is given by

$\text{Temperature}\text{lost}\text{by}\text{Fe}\text{block}\text{=}\text{Temperature}\text{gained}\text{by}\text{Al}\text{block}$

So the heat term is given by

$\begin{array}{l}\text{q}\text{=}\text{s}\text{×}\text{m}\text{×}\text{Δt}\\ \text{here,}\\ \text{ s}\text{-}\text{specific heat}\\ \text{m}\text{-}\text{mass}\text{and}\text{Δt}\text{-}\text{temperature}\text{change}\text{.}\end{array}$

$\left(\text{0}\text{.901}\text{J/g}\cdot {}^{\text{o}}\text{C}\right)\left(\text{20}\text{g}\right)\left({\text{t}}_{\text{f}}\text{-}\text{45}{}^{\text{o}}\text{C}\right)\text{=}\text{-}\left(\text{0}\text{.449}\text{J/g}\cdot {}^{\text{o}}\text{C}\right)\left(\text{20}\text{g}\right)\left({\text{t}}_{\text{f}}\text{-}\text{50}{}^{\text{o}}\text{C}\right)$

Simply the above equation and cancel the units

$\begin{array}{l}\left(\text{0}\text{.901}\right)\left({\text{t}}_{\text{f}}\text{-}\text{45}{}^{\text{o}}\text{C}\right)\text{=}\left(\text{0}\text{.449}\right)\left(\text{50}{}^{\text{o}}\text{C}\text{-}{\text{t}}_{\text{f}}\right)\\ \text{0}\text{.901}{\text{t}}_{\text{f}}\text{-}\text{40}\text{.545}\text{=}\text{22}\text{.45}\text{-}\text{0}\text{.449}{\text{t}}_{\text{f}}\\ \text{1}\text{.350}{\text{t}}_{\text{f}}\text{=}\text{62}\text{.995}\\ {\text{t}}_{\text{f}}\text{=}\text{46}{\text{.7}}^{\text{o}}\text{C}\end{array}$

The temperature of the both blocks will be $\text{46}\text{.7}{}^{\text{o}}\text{C}$ when ${\text{q}}_{\text{sys}}$ of iron block equals to the ${\text{q}}_{\text{sys}}$ of the aluminum block.