You lower the temperature of a sample of liquid carbon disulfide from 80.1 °C until its volume contracts by 0.543% of its initial value. What is the final temperature of the substance? The coefficient of volume expansion for carbon disulfide is 1.15 x 10-3 ('C)-'.
Energy transfer
The flow of energy from one region to another region is referred to as energy transfer. Since energy is quantitative; it must be transferred to a body or a material to work or to heat the system.
Molar Specific Heat
Heat capacity is the amount of heat energy absorbed or released by a chemical substance per the change in temperature of that substance. The change in heat is also called enthalpy. The SI unit of heat capacity is Joules per Kelvin, which is (J K-1)
Thermal Properties of Matter
Thermal energy is described as one of the form of heat energy which flows from one body of higher temperature to the other with the lower temperature when these two bodies are placed in contact to each other. Heat is described as the form of energy which is transferred between the two systems or in between the systems and their surrounding by the virtue of difference in temperature. Calorimetry is that branch of science which helps in measuring the changes which are taking place in the heat energy of a given body.
![**Problem Statement:**
You lower the temperature of a sample of liquid carbon disulfide from 80.1 °C until its volume contracts by 0.543% of its initial value. What is the final temperature of the substance? The coefficient of volume expansion for carbon disulfide is \( 1.15 \times 10^{-3} \, (\text{°C})^{-1} \).
**Solution:**
To solve for the final temperature, we can use the formula for volume expansion:
\[
\Delta V = V_i \times \beta \times \Delta T
\]
Where:
- \(\Delta V\) is the change in volume (given as a percentage of the initial volume, so \(\Delta V = -0.543\%\))
- \(V_i\) is the initial volume
- \(\beta\) is the coefficient of volume expansion (\(1.15 \times 10^{-3} \, (\text{°C})^{-1}\))
- \(\Delta T\) is the change in temperature
Rearranging the formula to solve for \(\Delta T\):
\[
\Delta T = \frac{\Delta V}{V_i \times \beta}
\]
Given that \(\Delta V / V_i = -0.00543\) (since it's a decrease),
\[
\Delta T = \frac{-0.00543}{1.15 \times 10^{-3}}
\]
Calculate \(\Delta T\) and then determine the final temperature by subtracting \(\Delta T\) from the initial temperature, 80.1 °C.
**Final Result:**
\[
\text{temperature:} \ \text{[Calculated Final Temperature]} \, °C
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