Chapter 10, Problem 5SP

(a)

To determine

The change in internal energy of the water in joules.

Answer to Problem 5SP

The change in internal energy of the water in joules is $2038\text{J}$.

Explanation of Solution

Given info: The amount of water in the beaker is $400\text{g}$ and the work done on water by stirring is $1200\text{J}$ and $200\text{cal}$ of heat added to it from a hot plate.

Work is done on the system, thus the work done will be negative.

Write the expression of first Law of Thermodynamics

$\begin{array}{c}\Delta U=Q-\left(-W\right)\\ =Q+W\end{array}$

Here,

$\Delta U$ is the change in internal energy

$Q$ is the heat transferred

$W$ is the work done

One calorie is equal to $4.19\text{J}$ thus, $200\text{cal}$ is equal to $838\text{J}$.

Substitute $1200\text{J}$ for $W$, and $838\text{J}$ for $Q$ in the above equation

$\begin{array}{c}\Delta U=1200\text{J+}838\text{J}\\ =2038\text{J}\end{array}$

Conclusion:

The change in internal energy of the water in joules is $2038\text{J}$.

(b)

To determine

The change in the internal energy of the water in calories.

Answer to Problem 5SP

The change in the internal energy of the water in calories is $486\text{cal}$.

Explanation of Solution

Given info:

Write the expression of first Law of Thermodynamics

$\begin{array}{c}\Delta U=Q-\left(-W\right)\\ =Q+W\end{array}$

Here,

$\Delta U$ is the change in internal energy

$Q$ is the heat transferred

$W$ is the work done

One calorie is equal to $4.19\text{J}$ thus, $1200\text{J}$ is equal to $286.4\text{cal}$.

Substitute $\text{286}\text{.4}\text{cal}$ for $W$, and $200\text{cal}$ for $Q$ in the above equation

$\begin{array}{c}\Delta U=286.4\text{cal+200}\text{cal}\\ =486.4\text{cal}\\ \approx 486\text{cal}\end{array}$

Conclusion:

The change in the internal energy of the water in calorie is $486\text{cal}$.

(c)

To determine

The temperature changes of the water.

Answer to Problem 5SP

The temperature change of the water is $1.2°\text{C}$.

Explanation of Solution

Write the expression for temperature change in terms of specific heat capacity

$\Delta T=\frac{Q}{mc}$

Here,

$Q$ is the heat

$m$ is the mass

$c$ is the specific heat capacity of ice

$\Delta T$ is the change in temperature

Substitute $400\text{g}$ for $m$, $\text{1 cal/g}\cdot °\text{C}$ for $c$ and $486.4\text{cal}$ for $Q$ in equation (1)

$\begin{array}{c}\Delta T=\frac{486.4\text{cal}}{400\text{g}\times \text{1 cal/g}\cdot °\text{C}}\\ =1.2°\text{C}\end{array}$

Conclusion:

The temperature change of the water is $1.2°\text{C}$.

(d)

To determine

To explain whether the answers in the first three parts will differ if the work done is $200\text{J}$ and heat added is $1200\text{cal}$.

Answer to Problem 5SP

Yes, Internal energy and temperature change are depending on the values of heat added or removed and work done, the answer will be different for each values of heat added and work done.

Explanation of Solution

Write the expression of first Law of Thermodynamics

$\begin{array}{c}\Delta U=Q-\left(-W\right)\\ =Q+W\end{array}$

Here,

$\Delta U$ is the change in internal energy

$Q$ is the heat transferred

$W$ is the work done

One calorie is equal to $4.19\text{J}$ thus, $1200\text{cal}$ is equal to $5028\text{J}$.

Substitute $200\text{J}$ for $W$, and $5028\text{J}$ for $Q$ in the above equation

$\begin{array}{c}\Delta U=200\text{J+}5028\text{J}\\ =5228\text{J}\\ \text{=}\frac{5228\text{J}}{4.19}\\ =1247.7\text{cal}\end{array}$

It is same as $1247.73\text{cal}$.

Write the expression for temperature change in terms of specific heat capacity

$\Delta T=\frac{Q}{mc}$ (1)

Here,

$Q$ is the heat

$m$ is the mass

$c$ is the specific heat capacity of ice

$\Delta T$ is the change in temperature

Substitute $400\text{g}$ for $m$ , $\text{1 cal/g}\cdot °\text{C}$ for $c$  and $1247.7\text{cal}$ for $Q$ in equation (1)

$\begin{array}{c}\Delta T=\frac{1247.7\text{cal}}{400\text{g}\times \text{1 cal/g}\cdot °\text{C}}\\ =3.12°\text{C}\end{array}$

All the three parts are giving different answers for $200\text{J}$ work done and $1200\text{cal}$ of heat added.

Conclusion:

Since internal energy and temperature change are depending on the values of heat added or removed and work done, the answer will be different for each values of heat added and work done.