QuestionFind $\Delta W$ and $\Delta U$ for a $6 \mathrm{~cm}$ iron cube heated from $20^{\circ} \mathrm{C}$ to $300^{\circ} \mathrm{C}$ .

Studdy Solution

STEP 1

Assumptions1. The initial temperature is $20^{\circ} \mathrm{C}$ . The final temperature is $300^{\circ} \mathrm{C}$
3. The specific heat capacity of iron is $0.1 \mathrm{cal}^{\mathrm{cal}} / \mathrm{g}^{\circ} \mathrm{C}$
4. The volume coefficient of thermal expansion for iron is $3.6 \times10^{-5} /{ }^{\Phi_{C}} \mathrm{C}$
5. The mass of the iron cube is $1700 \mathrm{~g}$

STEP 2

First, we need to find the change in internal energy $\Delta \mathrm{}$ , which can be calculated using the formula $\Delta \mathrm{} = m \cdot c \cdot \Delta$ where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta$ is the change in temperature.

STEP 3

Now, plug in the given values for the mass, specific heat capacity, and change in temperature to calculate the change in internal energy.
$\Delta \mathrm{} =1700 \mathrm{~g} \cdot0.1 \mathrm{cal}^{\mathrm{cal}} / \mathrm{g}^{\circ} \mathrm{C} \cdot (300^{\circ} \mathrm{C} -20^{\circ} \mathrm{C})$

STEP 4

Calculate the change in internal energy.
$\Delta \mathrm{} =1700 \mathrm{~g} \cdot0.1 \mathrm{cal}^{\mathrm{cal}} / \mathrm{g}^{\circ} \mathrm{C} \cdot280^{\circ} \mathrm{C} =47600 \mathrm{cal}$

STEP 5

Next, we need to find the change in work $\Delta \mathrm{W}$ , which can be calculated using the formula $\Delta \mathrm{W} = p \cdot \Delta V$ where $p$ is the pressure and $\Delta V$ is the change in volume. However, we first need to calculate $\Delta V$ .

STEP 6

The change in volume $\Delta V$ can be calculated using the formula $\Delta V = V \cdot \beta \cdot \Delta$ where $V$ is the initial volume, $\beta$ is the volume coefficient of thermal expansion, and $\Delta$ is the change in temperature.

STEP 7

The initial volume $V$ of the cube can be calculated using the formula $V = a^3$ where $a$ is the side length of the cube. Given that the side length is $6.0 \mathrm{~cm}$ , we have $V = (6.0 \mathrm{~cm})^3 =216 \mathrm{~cm}^3 =0.000216 \mathrm{~m}^3$

STEP 8

Now, plug in the given values for the initial volume, volume coefficient of thermal expansion, and change in temperature to calculate the change in volume.
$\Delta V =0.000216 \mathrm{~m}^3 \cdot3.6 \times10^{-5} /{ }^{\Phi_{C}} \mathrm{C} \cdot280^{\circ} \mathrm{C}$

STEP 9

Calculate the change in volume.
$\Delta V =.000216 \mathrm{~m}^3 \cdot3.6 \times^{-5} /{ }^{\Phi_{C}} \mathrm{C} \cdot280^{\circ} \mathrm{C} =2.18 \times^{-6} \mathrm{~m}^{3}$ So, the correct option is A $2.18 \times^{-6} \mathrm{~m}^{3}$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms