Chapter 5, Problem 44Q

To determine

To rank the final temperature of water after adding different containers.

Answer to Problem 44Q

The final temperature of water according to different added container is ranked from largest to smallest temperature.

$T_3>T_2>T_6>T_4>T_5>T_1$.

Explanation of Solution

Container A:

Given info:

The mass of water is $1500g$.

The temperature of water is $24°C$.

The mass of the container A is $100g$ and temperature is $30°C$.

The specific heat value for copper is $0.385J{g}^{-1}C{°}^{-1}$.

The specific heat value for water is $4.184J{g}^{-1}C{°}^{-1}$.

Formula used:

Formula to find the Final equilibrium temperature:

As amount of heat lost by copper is equal to the amount of heat gained by water,

$m_c\left(T_c-T\right)s_{copper}=m_w(T-T_w-)s_{water}$

Here, $m_c$ is the mass of copper, $T_c$ is the initial temperature of the copper container A, $T$ is the equilibrium temperature, $s_{copper}$ is the specific heat capacity of the copper. $m_w$ is the mass of water, $T_w$ is the initial temperature of the water, $T$ is the equilibrium temperature, $s_{water}$ is the specific heat capacity of the water.

Calculation:

Substitute the given values to find the final equilibrium temperature,

$\begin{array}{l}100g(30°C-T_1)0.385J{g}^{-1}C{°}^{-1}=1500g\left(T_1-24°C\right)4.184J{g}^{-1}C{°}^{-1}\\ 38.5(30°C-T_1)=6276\left(T_1-24°C\right)\\ (30°C-T_1)=163.013\left(T_1-24°C\right)\end{array}$

After simplifying we get,

$T_1=24.04C°$

Container B:

Given info:

The mass of water is $1500g$.

The temperature of water is $24°C$.

The mass of the container B is $200g$ and temperature is $60°C$.

The specific heat value for copper is $0.385J{g}^{-1}C{°}^{-1}$.

The specific heat value for water is $4.184J{g}^{-1}C{°}^{-1}$.

Formula used:

Formula to find the Final equilibrium temperature:

As amount of heat lost by copper is equal to the amount of heat gained by water,

$m_c\left(T_c-T\right)s_{copper}=m_w(T-T_w-)s_{water}$

Here, $m_c$ is the mass of copper, $T_c$ is the initial temperature of the copper container A, $T$ is the equilibrium temperature, $s_{copper}$ is the specific heat capacity of the copper. $m_w$ is the mass of water, $T_w$ is the initial temperature of the water, $T$ is the equilibrium temperature, $s_{water}$ is the specific heat capacity of the water.

Calculation:

Substitute the given values to find the final equilibrium temperature,

$\begin{array}{l}200g(60°C-T_2)0.385J{g}^{-1}C{°}^{-1}=1500g\left(T_2-24°C\right)4.184J{g}^{-1}C{°}^{-1}\\ 38.5(60°C-T_2)=6276\left(T_2-24°C\right)\\ (60°C-T_2)=81.51\left(T_2-24°C\right)\end{array}$

After simplifying we get,

$T_2=24.43C°$

Container C:

Given info:

The mass of water is $1500g$.

The temperature of water is $24°C$.

The mass of the container C is $300g$ and temperature is $90°C$.

The specific heat value for copper is $0.385J{g}^{-1}C{°}^{-1}$.

The specific heat value for water is $4.184J{g}^{-1}C{°}^{-1}$.

Formula used:

Formula to find the Final equilibrium temperature:

As amount of heat lost by copper is equal to the amount of heat gained by water,

$m_c\left(T_c-T\right)s_{copper}=m_w(T-T_w-)s_{water}$

Here, $m_c$ is the mass of copper, $T_c$ is the initial temperature of the copper container A, $T$ is the equilibrium temperature, $s_{copper}$ is the specific heat capacity of the copper. $m_w$ is the mass of water, $T_w$ is the initial temperature of the water, $T$ is the equilibrium temperature, $s_{water}$ is the specific heat capacity of the water.

Calculation:

Substitute the given values to find the final equilibrium temperature,

$\begin{array}{l}300g(90°C-T_3)0.385J{g}^{-1}C{°}^{-1}=1500g\left(T_3-24°C\right)4.184J{g}^{-1}C{°}^{-1}\\ 115.5(90°C-T_3)=6276\left(T_3-24°C\right)\\ (90°C-T_3)=54.34\left(T_3-24°C\right)\end{array}$

After simplifying we get,

$T_3=25.19C°$

Container D:

Given info:

The mass of water is $1500g$.

The temperature of water is $24°C$.

The mass of the container D is $200g$ and temperature is $15°C$.

The specific heat value for copper is $0.385J{g}^{-1}C{°}^{-1}$.

The specific heat value for water is $4.184J{g}^{-1}C{°}^{-1}$.

Formula used:

Formula to find the Final equilibrium temperature:

As amount of heat lost by copper is equal to the amount of heat gained by water,

$m_c\left(T_c-T\right)s_{copper}=m_w(T-T_w-)s_{water}$

Here, $m_c$ is the mass of copper, $T_c$ is the initial temperature of the copper container A, $T$ is the equilibrium temperature, $s_{copper}$ is the specific heat capacity of the copper. $m_w$ is the mass of water, $T_w$ is the initial temperature of the water, $T$ is the equilibrium temperature, $s_{water}$ is the specific heat capacity of the water.

Calculation:

Substitute the given values to find the final equilibrium temperature,

$\begin{array}{l}200g(15°C-T_4)0.385J{g}^{-1}C{°}^{-1}=1500g\left(T_4-24°C\right)4.184J{g}^{-1}C{°}^{-1}\\ 77(15°C-T_4)=6276\left(T_4-24°C\right)\\ (15°C-T_4)=81.51\left(T_4-24°C\right)\end{array}$

After simplifying we get,

$T_4=24.183C°$

Container E:

Given info:

The mass of water is $1500g$.

The temperature of water is $24°C$.

The mass of the container E is $300g$ and temperature is $30°C$.

The specific heat value for copper is $0.385J{g}^{-1}C{°}^{-1}$.

The specific heat value for water is $4.184J{g}^{-1}C{°}^{-1}$.

Formula used:

Formula to find the Final equilibrium temperature:

As amount of heat lost by copper is equal to the amount of heat gained by water,

$m_c\left(T_c-T\right)s_{copper}=m_w(T-T_w-)s_{water}$

Here, $m_c$ is the mass of copper, $T_c$ is the initial temperature of the copper container A, $T$ is the equilibrium temperature, $s_{copper}$ is the specific heat capacity of the copper. $m_w$ is the mass of water, $T_w$ is the initial temperature of the water, $T$ is the equilibrium temperature, $s_{water}$ is the specific heat capacity of the water.

Calculation:

Substitute the given values to find the final equilibrium temperature,

$\begin{array}{l}300g(30°C-T_5)0.385J{g}^{-1}C{°}^{-1}=1500g\left(T_5-24°C\right)4.184J{g}^{-1}C{°}^{-1}\\ 115.5(30°C-T_5)=6276\left(T_5-24°C\right)\\ (30°C-T_5)=54.34\left(T_5-24°C\right)\end{array}$

After simplifying we get,

$T_5=24.11C°$

Container F:

Given info:

The mass of water is $1500g$.

The temperature of water is $24°C$.

The mass of the container F is $100g$ and temperature is $60°C$.

The specific heat value for copper is $0.385J{g}^{-1}C{°}^{-1}$.

The specific heat value for water is $4.184J{g}^{-1}C{°}^{-1}$.

Formula used:

Formula to find the Final equilibrium temperature:

As amount of heat lost by copper is equal to the amount of heat gained by water,

$m_c\left(T_c-T\right)s_{copper}=m_w(T-T_w-)s_{water}$

Here, $m_c$ is the mass of copper, $T_c$ is the initial temperature of the copper container A, $T$ is the equilibrium temperature, $s_{copper}$ is the specific heat capacity of the copper. $m_w$ is the mass of water, $T_w$ is the initial temperature of the water, $T$ is the equilibrium temperature, $s_{water}$ is the specific heat capacity of the water.

Calculation:

Substitute the given values to find the final equilibrium temperature,

$\begin{array}{l}100g(60°C-T_6)0.385J{g}^{-1}C{°}^{-1}=1500g\left(T_6-24°C\right)4.184J{g}^{-1}C{°}^{-1}\\ 38.5(60°C-T_6)=6276\left(T_6-24°C\right)\\ (60°C-T_6)=163.013\left(T_6-24°C\right)\end{array}$

After simplifying we get,

$T_6=24.22C°$

After calculation the final temperature of water according to different added container is ranked from largest to smallest temperature as,

$T_3>T_2>T_6>T_4>T_5>T_1$.

Conclusion:

Thus, the final temperature of water is ranked from largest to smallest temperature as,

$T_3>T_2>T_6>T_4>T_5>T_1$.